CBSE Case Study Questions for Class 10 Maths Coordinate Geometry Free PDF
Mere Bacchon, you must practice the CBSE Case Study Questions Class 10 Maths Coordinate Geometry in order to fully complete your preparation . They are very very important from exam point of view. These tricky Case Study Based Questions can act as a villain in your heroic exams!
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CBSE Case Study Questions for Class 10 Maths Coordinate Geometry PDF
Mcq set 1 -, mcq set 2 -, checkout our case study questions for other chapters.
- Chapter 5: Arithmetic Progressions Case Study Questions
- Chapter 6: Triangles Case Study Questions
- Chapter 8: Introduction to Trigonometry Case Study Questions
- Chapter 9: Some Applications of Trigonometry Case Study Questions
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Case Study Questions for Class 10 Maths Chapter 7 Coordinate Geometry
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Case Study Questions:
Question 1:
The top of a table is shown in the figure given below:
(i) The coordinates of the points H and G are respectively (a) (1, 5), (5, 1) (b) (0, 5), (5, 0) (c) (1, 5), (5, 0) (d) (5, 1), (1, 5)
(ii) The distance between the points A and B is (a) 4 units (b) 4 2 units (c) 16 units (d) 32 units
(iii) The coordinates of the mid point of line segment joining points M and Q are (a) (9, 3) (b) (5, 11) (c) (14, 14) (d) (7, 7)
(iv) Which among the following have same ordinate? (a) H and A (b) T and O (c) R and M (d) N and R
(v) If G is taken as the origin, and x, y axis put along GF and GB, then the point denoted by coordinate (4, 2) is (a) H (b) F (c) Q (d) R
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Chapter 1 Real Numbers Chapter 2 Polynomials Chapter 3 Pair of Linear Equations in Two Variables C hapter 4 Quadratic Equations Chapter 5 Arithmetic Progressions Chapter 6 Triangles Chapter 7 Coordinate Geometry Chapter 8 Introduction to Trigonometry Chapter 9 Some Applications of Trigonometry Chapter 10 Circles Chapter 11 Constructions Chapter 12 Areas Related to Circles Chapter 13 Surface Areas and Volumes Chapter 14 Statistics Chapter 15 Probability
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CBSE Class 10 Maths Case Study Questions PDF
Download Case Study Questions for Class 10 Mathematics to prepare for the upcoming CBSE Class 10 Final Exam. These Case Study and Passage Based questions are published by the experts of CBSE Experts for the students of CBSE Class 10 so that they can score 100% on Boards.
CBSE Class 10 Mathematics Exam 2024 will have a set of questions based on case studies in the form of MCQs. The CBSE Class 10 Mathematics Question Bank on Case Studies, provided in this article, can be very helpful to understand the new format of questions. Share this link with your friends.
Table of Contents
Chapterwise Case Study Questions for Class 10 Mathematics
Inboard exams, students will find the questions based on assertion and reasoning. Also, there will be a few questions based on case studies. In that, a paragraph will be given, and then the MCQ questions based on it will be asked.
The above Case studies for Class 10 Maths will help you to boost your scores as Case Study questions have been coming in your examinations. These CBSE Class 10 Mathematics Case Studies have been developed by experienced teachers of cbseexpert.com for the benefit of Class 10 students.
- Class 10th Science Case Study Questions
- Assertion and Reason Questions of Class 10th Science
- Assertion and Reason Questions of Class 10th Social Science
Class 10 Maths Syllabus 2024
Chapter-1 real numbers.
Starting with an introduction to real numbers, properties of real numbers, Euclid’s division lemma, fundamentals of arithmetic, Euclid’s division algorithm, revisiting irrational numbers, revisiting rational numbers and their decimal expansions followed by a bunch of problems for a thorough and better understanding.
Chapter-2 Polynomials
This chapter is quite important and marks securing topics in the syllabus. As this chapter is repeated almost every year, students find this a very easy and simple subject to understand. Topics like the geometrical meaning of the zeroes of a polynomial, the relationship between zeroes and coefficients of a polynomial, division algorithm for polynomials followed with exercises and solved examples for thorough understanding.
Chapter-3 Pair of Linear Equations in Two Variables
This chapter is very intriguing and the topics covered here are explained very clearly and perfectly using examples and exercises for each topic. Starting with the introduction, pair of linear equations in two variables, graphical method of solution of a pair of linear equations, algebraic methods of solving a pair of linear equations, substitution method, elimination method, cross-multiplication method, equations reducible to a pair of linear equations in two variables, etc are a few topics that are discussed in this chapter.
Chapter-4 Quadratic Equations
The Quadratic Equations chapter is a very important and high priority subject in terms of examination, and securing as well as the problems are very simple and easy. Problems like finding the value of X from a given equation, comparing and solving two equations to find X, Y values, proving the given equation is quadratic or not by knowing the highest power, from the given statement deriving the required quadratic equation, etc are few topics covered in this chapter and also an ample set of problems are provided for better practice purposes.
Chapter-5 Arithmetic Progressions
This chapter is another interesting and simpler topic where the problems here are mostly based on a single formula and the rest are derivations of the original one. Beginning with a basic brief introduction, definitions of arithmetic progressions, nth term of an AP, the sum of first n terms of an AP are a few important and priority topics covered under this chapter. Apart from that, there are many problems and exercises followed with each topic for good understanding.
Chapter-6 Triangles
This chapter Triangle is an interesting and easy chapter and students often like this very much and a securing unit as well. Here beginning with the introduction to triangles followed by other topics like similar figures, the similarity of triangles, criteria for similarity of triangles, areas of similar triangles, Pythagoras theorem, along with a page summary for revision purposes are discussed in this chapter with examples and exercises for practice purposes.
Chapter-7 Coordinate Geometry
Here starting with a general introduction, distance formula, section formula, area of the triangle are a few topics covered in this chapter followed with examples and exercises for better and thorough practice purposes.
Chapter-8 Introduction to Trigonometry
As trigonometry is a very important and vast subject, this topic is divided into two parts where one chapter is Introduction to Trigonometry and another part is Applications of Trigonometry. This Introduction to Trigonometry chapter is started with a general introduction, trigonometric ratios, trigonometric ratios of some specific angles, trigonometric ratios of complementary angles, trigonometric identities, etc are a few important topics covered in this chapter.
Chapter-9 Applications of Trigonometry
This chapter is the continuation of the previous chapter, where the various modeled applications are discussed here with examples and exercises for better understanding. Topics like heights and distances are covered here and at the end, a summary is provided with all the important and frequently used formulas used in this chapter for solving the problems.
Chapter-10 Circle
Beginning with the introduction to circles, tangent to a circle, several tangents from a point on a circle are some of the important topics covered in this chapter. This chapter being practical, there are an ample number of problems and solved examples for better understanding and practice purposes.
Chapter-11 Constructions
This chapter has more practical problems than theory-based definitions. Beginning with a general introduction to constructions, tools used, etc, the topics like division of a line segment, construction of tangents to a circle, and followed with few solved examples that help in solving the exercises provided after each topic.
Chapter-12 Areas related to Circles
This chapter problem is exclusively formula based wherein topics like perimeter and area of a circle- A Review, areas of sector and segment of a circle, areas of combinations of plane figures, and a page summary is provided just as a revision of the topics and formulas covered in the entire chapter and also there are many exercises and solved examples for practice purposes.
Chapter-13 Surface Areas and Volumes
Starting with the introduction, the surface area of a combination of solids, the volume of a combination of solids, conversion of solid from one shape to another, frustum of a cone, etc are to name a few topics explained in detail provided with a set of examples for a better comprehension of the concepts.
Chapter-14 Statistics
In this chapter starting with an introduction, topics like mean of grouped data, mode of grouped data, a median of grouped, graphical representation of cumulative frequency distribution are explained in detail with exercises for practice purposes. This chapter being a simple and easy subject, securing the marks is not difficult for students.
Chapter-15 Probability
Probability is another simple and important chapter in examination point of view and as seeking knowledge purposes as well. Beginning with an introduction to probability, an important topic called A theoretical approach is explained here. Since this chapter is one of the smallest in the syllabus and problems are also quite easy, students often like this chapter
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Class 10 Maths Chapter 7 Case Based Questions - Coordinate Geometry
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Study Case - 1
The class X students of a school in Rajinder Nagar have been allotted a rectangular plot of land for their gardening activity. Sapling of Mango are planted on the boundary at the distance of 1 m from each other.
Based on the above figure, answer the following questions:
Q4: What will be the coordinates of the vertices of ΔPQR if C is the origin? (a) (14, 3), (11, 2), (8, 6) (b) (15, 4), (12, 3), (9, 7) (c) (14, 2), (11, 1), (8, 5) (d) (15, 3), (12, 2), (9, 6) Ans: (d) Explanation: When C is taken as origin, we will take CB as X-axis and CD as Y-axis. Then, coordinates of points, P, Q and R are (15, 3), (12, 2)and (9, 6) respectively.
Q5: What are the coordinates of P if D is taken as the origin? (a) (−15, 5) (b) (15, 5) (c) (15, 7) (d) (15, 3) Ans: (a) Explanation: When D is taken as origin, we will take DA as negative X-axis and DC as positive Y-axis. Then coordinates of point P is (−15, 5).
Case Study - 2
Based on the above information, give the answer of the following questions:
Study Case - 3
Based on the above information give the answer of the following questions:
Q1: The coordinates of the point A are: (a) (4,9) (b) (5,9) (c) (92,9) (d) (4,8) Ans: (c) Explanation: As the distance of point A is 9/2 units from the Y-axis and 9 units from the X-axis, its coordinates are x = 9/2, y = 9 or (9/2, 9)
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Class 10th Maths - Coordinate Geometry Case Study Questions and Answers 2022 - 2023
Class 10th Maths - Coordinate Geometry Case Study Questions and Answers 2022 - 2023 Study Materials Sep-09 , 2022
QB365 provides a detailed and simple solution for every Possible Case Study Questions in Class 10th Maths Subject - Coordinate Geometry, CBSE. It will help Students to get more practice questions, Students can Practice these question papers in addition to score best marks.
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Coordinate geometry case study questions with answer key.
Final Semester - June 2015
(ii) The value of x + y is
(iii) Which of the following is true?
(iv) The ratio in which B divides AC is
(v) Which of the following equations is satisfied by the given points?
(ii) How far is the library from Shaguns house?
(iii) How far is the library from Alia's house?
(iv) Which of the following is true?
(v) How far is the school from Alia's house than Shaguns house?
| units | units | + )units | - ) units |
(ii) The value of x is equal to
(iii) If M is any point exactly in between city A and city B, then coordinates of M are
(iv) The ratio in which A divides the line segment joining the points O and M is
(v) If the person analyse the petrol at the point M(the mid point of AB), then what should be his decision?
(ii) The distance of the bus stand from the house is
(iii) If the grocery store and electrician's shop lie on a line, the ratio of distance of house from grocery store to that from electrician's shop, is
(iv) The ratio of distances of house from bus stand to food cart is
(v) The coordinates of positions of bus stand, grocery store, food cart and electrician's shop form a
| d) none of these |
| \(A\left(\frac{2}{3}, 0\right),\) | \((b) \left(0, \frac{2}{3}\right)\) | \((c) \left(0, \frac{4}{3}\right)\) | \((d) \left(\frac{4}{3}, 0\right)\) |
(ii) The centre of circle is the
(iii) The radius of the circle is
| \((a) \frac{4}{3} units\) | \((b) \frac{3}{2} units\) | \((c) \frac{2}{3} units\) | \((d) \frac{3}{4} units\) |
(iv) The area of the circle is
| \((a) 16 \pi^{2} sq. units\) | \((b) \frac{16}{9} \pi sq. units\) | \((c) \frac{4}{9} \pi^{2} sq. units\) | \((d) 4 \pi sq. units\) |
(v) If \(\left(1, \frac{\sqrt{7}}{3}\right)\) is one of the ends of a diameter, then its other end is
| \((a) \left(-1, \frac{\sqrt{7}}{3}\right)\) | \((b) \left(1,-\frac{\sqrt{7}}{3}\right)\) | \((c) \left(1, \frac{\sqrt{7}}{3}\right)\) | \((d) \left(-1,-\frac{\sqrt{7}}{3}\right)\) |
| \((a) \left(\frac{2}{3}, \frac{7}{3}\right)^{n}\) | \((b) \left(\frac{1}{3}, \frac{1}{3}\right)\) | \((c) \left(\frac{-2}{3}, \frac{7}{3}\right)\) | \((d) \left(\frac{7}{3}, \frac{2}{3}\right)\) |
(ii) If S be the mid-point of line joining P and Q, then coordinates of S are
(iii) If T be the mid-point of line joining Rand Q, then coordinates of T are
| \((a) \left(\frac{1}{2}, \frac{1}{2}\right)\) | \((b) \left(\frac{3}{2}, \frac{1}{2}\right)\) | \((c) \left(\frac{1}{2}, \frac{3}{2}\right)\) |
(iv) If Ube the mid-point of line joining Rand P, then coordinates of U are
| \((a) \left(-\frac{5}{2}, \frac{3}{2}\right)\) | \((b) \left(\frac{3}{2},-\frac{5}{2}\right)\) | \((c) \left(\frac{3}{2}, \frac{5}{2}\right)\) | \((d) \left(\frac{5}{2}, \frac{3}{2}\right)\) |
(v) The coordinates of centroid of \(\Delta\) STU are
| \((a) \left(\frac{2}{3}, \frac{7}{3}\right)\) | \((b) \left(\frac{1}{3}, \frac{1}{3}\right)\) | \((c) \left(-\frac{2}{3}, \frac{7}{3}\right)\) | \((d) \left(\frac{7}{3}, \frac{2}{3}\right)\) |
| km |
(ii) The distance between A and Cis
| km | km |
(iii) If it is assumed that both buses have same speed, then by which bus do you want to travel from A to B?
(iv) If the fare for first bus is Rs10/km, then what will be the fare for total journey by that bus?
(v) If the fare for second bus is Rs 15/km, then what will be the fare to reach to the destination by this bus?
(ii) Perimeter of the region covered by group A is
| units | units | + \(\sqrt{13}\)units |
(iii) If the coordinates of region covered by group B, taken in the same order forms a quadrilateral, then the length of each of its diagonals is
| units, \(2\sqrt{2}\) units | units, \(\sqrt{2}\) units |
| units, \(2\sqrt{2}\) units |
(iv) If region covered by group B forms a rhombus, where the coordinates are taken in given order, then the perimeter of this region is
| units | units | units | units |
(v) The coordinates of the point which divides the join of points P(x 1 , y 1 ) and Q(x 2 , y 2 ) internally in the ratio m: n is
| \(\text { (a) }\left(\frac{m x_{2}+n y_{2}}{m+n}, \frac{m x_{1}+n y_{1}}{m+n}\right)\) | \(\text { (b) }\left(\frac{m x_{1}+n y_{1}}{m+n}, \frac{m x_{2}+n y_{2}}{m+n}\right)\) |
| \(\text { (c) }\left(\frac{m x_{2}+n x_{1}}{m+n}, \frac{m y_{2}+n y_{1}}{m+n}\right)\) |
| units | units | units |
(ii) If an invigilator at the point I, lying on the straight line joining Band C such that it divides the distance between them in the ratio of 1 : 2. Then coordinates of I are
(iii) The mid-point of the line segment joining A and C is
| ne of these |
(iv) The ratio in which B divides the line segment joining A and C is
| ne of these |
(v) The points A, Band C lie on
| |
(ii) Length of BC is
(iii) Length of FG is
(iv) The mid-point of FG lies on line represented by
(v) The perimeter of its trunk LMPN is
| units | units |
(ii) What are the coordinates of U?
(iii) The distance between the points P and Q is
(iv) If a point A(x, y) is equidistant from Rand T, then
(v) Perimeter of image of a rectangular face is
(ii) The mid point of the line segment joining A and C is
(iii) Which of the following is near to A?
(iv) Which of the following is equidistant from Band D?
(v) If Q is considered as origin, then the coordinates of D will be
(ii) Distance of the point G from the y-axis is
(iii) Length of side HG =
(iv) The length of diagonal FD and the value of x, respectively are
| units,5 | units,9 | units,9 |
(v) If Q is considered as origin, then the coordinates of mid-point of BC are
(ii) What is the distance between A and C?
| units | units | units |
(iii) Which of the following distance is least?
(iv) Which of the following is the best route to go to the office?
(v) If Alok and Deepika planned to meet at a mall situated at a point D represented by the mid-point of the line joining the points Band C, then find the coordinates of D.
(b) At what distance Preet posted the green flag from the starting point of eighth line?
(c) What is the distance between both the flags?
(d) If Rashmi has to post a blue flag exactly halfway between the line segments joining the two flags, where should she post her flag?
(e) If Shweta has to post a white flag exactly halfway between the line segments joining A and red flag, where should she post her flag?
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CBSE Class 10 Maths Competency-Based Questions With Answer Key 2024-25: Chapter 7 Coordinate Geometry Free PDF Download
Cbse class 10 maths chapter 7 practice questions 2025: cbse class 10 students download the maths chapter 7 coordinate geometry competency-focused practice questions along with answer key for 2024-25. .
What are Competency-Focused Practice Questions?
Download volume 3 and 4 of cbse class 10 maths chapter 7 coordinate geometry, cbse class 10 maths chapter 7 competency based-questions .
| Shown below are 2 identical rectangles such that their breadth is half their length. |
| Shown below is a quarter of a circle with centre at (0, 0). An arbitrary point R lies on the boundary of the quadrant. |
| A circle has its centre at the origin. The radius of the circle is 5 units. |
| Arshad was eating chips while working with graph paper. One chip fell on his graph paper as shown below. |
| Shown below is the map of India put on a coordinate plane where each small square represents 1 sq unit. (Check pdf for image). |
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1. Mathematics is a subject that demands regular practice. It is important that students practice as much as they can. Practice the questions given at the end of each chapter. Note the points where you get stuck during solving questions.
2. Students should prepare an effective study plan and give time to all topics on the basis of their strengths and weaknesses. Don’t let your bias towards your favourite chapter affect the time-table. The schedule should also have time for revision.
3. Students should solve the problems from CBSE Class 10 subject textbook to clear the concepts and strengthen understanding. While solving problems, students should make it a habit to show rough work clearly which will help them in the examination as well.
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Case Study Class 10 Maths Questions
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Now, CBSE will ask only subjective questions in class 10 Maths case studies. But if you search over the internet or even check many books, you will get only MCQs in the class 10 Maths case study in the session 2022-23. It is not the correct pattern. Just beware of such misleading websites and books.
We advise you to visit CBSE official website ( cbseacademic.nic.in ) and go through class 10 model question papers . You will find that CBSE is asking only subjective questions under case study in class 10 Maths. We at myCBSEguide helping CBSE students for the past 15 years and are committed to providing the most authentic study material to our students.
Here, myCBSEguide is the only application that has the most relevant and updated study material for CBSE students as per the official curriculum document 2022 – 2023. You can download updated sample papers for class 10 maths .
First of all, we would like to clarify that class 10 maths case study questions are subjective and CBSE will not ask multiple-choice questions in case studies. So, you must download the myCBSEguide app to get updated model question papers having new pattern subjective case study questions for class 10 the mathematics year 2022-23.
Class 10 Maths has the following chapters.
- Real Numbers Case Study Question
- Polynomials Case Study Question
- Pair of Linear Equations in Two Variables Case Study Question
- Quadratic Equations Case Study Question
- Arithmetic Progressions Case Study Question
- Triangles Case Study Question
- Coordinate Geometry Case Study Question
- Introduction to Trigonometry Case Study Question
- Some Applications of Trigonometry Case Study Question
- Circles Case Study Question
- Area Related to Circles Case Study Question
- Surface Areas and Volumes Case Study Question
- Statistics Case Study Question
- Probability Case Study Question
Format of Maths Case-Based Questions
CBSE Class 10 Maths Case Study Questions will have one passage and four questions. As you know, CBSE has introduced Case Study Questions in class 10 and class 12 this year, the annual examination will have case-based questions in almost all major subjects. This article will help you to find sample questions based on case studies and model question papers for CBSE class 10 Board Exams.
Maths Case Study Question Paper 2023
Here is the marks distribution of the CBSE class 10 maths board exam question paper. CBSE may ask case study questions from any of the following chapters. However, Mensuration, statistics, probability and Algebra are some important chapters in this regard.
| I | NUMBER SYSTEMS | 06 |
| II | ALGEBRA | 20 |
| III | COORDINATE GEOMETRY | 06 |
| IV | GEOMETRY | 15 |
| V | TRIGONOMETRY | 12 |
| V | MENSURATION | 10 |
| VI | STATISTICS & PROBABILITY | 11 |
Case Study Question in Mathematics
Here are some examples of case study-based questions for class 10 Mathematics. To get more questions and model question papers for the 2021 examination, download myCBSEguide Mobile App .
Case Study Question – 1
In the month of April to June 2022, the exports of passenger cars from India increased by 26% in the corresponding quarter of 2021–22, as per a report. A car manufacturing company planned to produce 1800 cars in 4th year and 2600 cars in 8th year. Assuming that the production increases uniformly by a fixed number every year.
- Find the production in the 1 st year.
- Find the production in the 12 th year.
- Find the total production in first 10 years. OR In which year the total production will reach to 15000 cars?
Case Study Question – 2
In a GPS, The lines that run east-west are known as lines of latitude, and the lines running north-south are known as lines of longitude. The latitude and the longitude of a place are its coordinates and the distance formula is used to find the distance between two places. The distance between two parallel lines is approximately 150 km. A family from Uttar Pradesh planned a round trip from Lucknow (L) to Puri (P) via Bhuj (B) and Nashik (N) as shown in the given figure below.
- Find the distance between Lucknow (L) to Bhuj(B).
- If Kota (K), internally divide the line segment joining Lucknow (L) to Bhuj (B) into 3 : 2 then find the coordinate of Kota (K).
- Name the type of triangle formed by the places Lucknow (L), Nashik (N) and Puri (P) OR Find a place (point) on the longitude (y-axis) which is equidistant from the points Lucknow (L) and Puri (P).
Case Study Question – 3
- Find the distance PA.
- Find the distance PB
- Find the width AB of the river. OR Find the height BQ if the angle of the elevation from P to Q be 30 o .
Case Study Question – 4
- What is the length of the line segment joining points B and F?
- The centre ‘Z’ of the figure will be the point of intersection of the diagonals of quadrilateral WXOP. Then what are the coordinates of Z?
- What are the coordinates of the point on y axis equidistant from A and G? OR What is the area of area of Trapezium AFGH?
Case Study Question – 5
The school auditorium was to be constructed to accommodate at least 1500 people. The chairs are to be placed in concentric circular arrangement in such a way that each succeeding circular row has 10 seats more than the previous one.
- If the first circular row has 30 seats, how many seats will be there in the 10th row?
- For 1500 seats in the auditorium, how many rows need to be there? OR If 1500 seats are to be arranged in the auditorium, how many seats are still left to be put after 10 th row?
- If there were 17 rows in the auditorium, how many seats will be there in the middle row?
Case Study Question – 6
- Draw a neat labelled figure to show the above situation diagrammatically.
- What is the speed of the plane in km/hr.
More Case Study Questions
We have class 10 maths case study questions in every chapter. You can download them as PDFs from the myCBSEguide App or from our free student dashboard .
As you know CBSE has reduced the syllabus this year, you should be careful while downloading these case study questions from the internet. You may get outdated or irrelevant questions there. It will not only be a waste of time but also lead to confusion.
Here, myCBSEguide is the most authentic learning app for CBSE students that is providing you up to date study material. You can download the myCBSEguide app and get access to 100+ case study questions for class 10 Maths.
How to Solve Case-Based Questions?
Questions based on a given case study are normally taken from real-life situations. These are certainly related to the concepts provided in the textbook but the plot of the question is always based on a day-to-day life problem. There will be all subjective-type questions in the case study. You should answer the case-based questions to the point.
What are Class 10 competency-based questions?
Competency-based questions are questions that are based on real-life situations. Case study questions are a type of competency-based questions. There may be multiple ways to assess the competencies. The case study is assumed to be one of the best methods to evaluate competencies. In class 10 maths, you will find 1-2 case study questions. We advise you to read the passage carefully before answering the questions.
Case Study Questions in Maths Question Paper
CBSE has released new model question papers for annual examinations. myCBSEguide App has also created many model papers based on the new format (reduced syllabus) for the current session and uploaded them to myCBSEguide App. We advise all the students to download the myCBSEguide app and practice case study questions for class 10 maths as much as possible.
Case Studies on CBSE’s Official Website
CBSE has uploaded many case study questions on class 10 maths. You can download them from CBSE Official Website for free. Here you will find around 40-50 case study questions in PDF format for CBSE 10th class.
10 Maths Case Studies in myCBSEguide App
You can also download chapter-wise case study questions for class 10 maths from the myCBSEguide app. These class 10 case-based questions are prepared by our team of expert teachers. We have kept the new reduced syllabus in mind while creating these case-based questions. So, you will get the updated questions only.
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Important Questions for Class 10 Maths Chapter 7 Coordinate Geometry
Chapter 7 coordinate geometry important questions for cbse class 10 maths board exams.
Important Questions for Chapter 7 Coordinate Geometry Class 10 Maths
Coordinate geometry class 10 maths important questions very short answer (1 mark).
Since it is equidistant from the points A(-2, 0) and B(6, 0) then
⇒ AP 2 = BP 2
Using distance formula we have,
[x-(2)] 2 + (0-0) 2 = (x+6) 2 + (0-0) 2
⇒ (x+2) 2 = (x+6) 2
⇒ x 2 + 4x + x = x 2 + 12x + 36
⇒ 8x = -32
⇒ x = -4
Hence, required point P is (-4, 0).
3. Find the distance of a point P(x,y) from the origin.
Distance between origin (0, 0) and point P(x,y) is
When the points are collinear, x 1 (y 2 – y 3 ) + x 2 (y 3 – y 1 ) + x 3 (y 1 – y 2 ) = 0 ⇒ x(-4 – (-5)) + (-3)(-5 – 2) + 7(2 – (-4)) = 0 ⇒ x(1) + 21 + 42 = 0 ⇒ x + 63 = 0 ∴ x = -63
7. For what value of k will k + 9, 2k – 1 and 2k + 7 are the consecutive terms of an A.P.?
As we know, a 2 – a 1 = a 3 – a 2 2k – 1 – (k + 9) = 2k + 7 – (2k – 1) ⇒ 2k – 1 – k – 9 = 2k + 7 – 2k + 1 ⇒ k – 10 = 8 ∴ k = 8 + 10 = 18
8. ABCD is a rectangle whose three vertices are B(4, 0), C(4, 3) and D(0, 3). Calculate the length of one of its diagonals.
AB = 4 units BC = 3 units AC 2 = AB 2 + BC 2 …(Pythagoras’ theorem) = (4) 2 + (3) 2 = 16 + 9 = 25 ∴ AC = 5 cm
Coordinate Geometry Class 10 Maths Important Questions Short Answer-I (2 Marks)
Let the point P be (2y, y). Since PQ = PR, we have
Squaring both sides,
(2y-2) 2 + (y+5) 2 = (2y+3) 2 + (y-6) 2
⇒ -8y + 4 + 10y + 25 = 12y + 9 -12y + 36
⇒ 2y + 29 = 45
⇒ y = 8
Hence, the coordinates of point P are (16,8).
When points are collinear, ∴ Area of ∆ABC = 0 = (x 1 (y 2 – y 3 ) + x 2 (y 3 – y 1 ) + x 3 (y 1 – y 2 )) = 0 = x (7 – 5) – 5 (5 – y) -4 (y – 7) = 0 = 2x – 25 + 5y – 4y + 28 = 0 ∴ 2x + y + 3 = 0 is the required relation.
13. Find a relation between x and y such that the point P(x, y) is equidistant from the points A (2, 5) and B (-3, 7).
Let P (x, y) be equidistant from the points A (2, 5) and B (-3, 7). ∴ AP = BP …(Given) AP 2 = BP 2 …(Squaring both sides) (x – 2) 2 + (y – 5) 2 = (x + 3) 2 + (y – 7) 2 ⇒ x 2 – 4x + 4 + y 2 – 10y + 25 ⇒ x 2 + 6x + 9 + y 2 – 14y + 49 ⇒ -4x – 10y – 6x + 14y = 9 +49 – 4 – 25 ⇒ -10x + 4y = 29 ∴ 10x + 29 = 4y is the required relation.
14. The points A(4,7), B(p,3) and C (7,3) are the vertices of a right triangle, right-angled at B. Find the value of p.
As per question, triangle is shown below. Here ABC is a right angle triangle,
AB 2 + BC 2 = AC 2
(p-4) 2 + (3-7) 2 + (7-p) 2 + (3-3) 2 = (7-4) 2 + (3-4) 2
⇒ (p-4) 2 + (-4) 2 + (7-p) 2 + 0 = 3 2 + (-4) 2
⇒ p 2 - 8p + 16 + 16 + 49 + p 2 - 14p = 9 + 16
⇒ 2p 2 - 22p + 81 = 25
⇒ 2p 2 - 22p + 56 = 0
⇒ p 2 - 11p + 28 = 0
⇒ (p-4) (p-7) = 0
⇒ p = 7 or 4
15. If A(4,3), B(-1,y) and C(3,4) are the vertices of a right triangle , ABC right angled at A, then find the value of y.
As per question, triangle is shown below.
AB 2 + AC 2 = BC 2
(4+1) 2 + (3-y) 2 + (4-3) 2 = (3+1) 2 + (4-y) 2
⇒ 5 2 + (3-y) 2 + (-1) 2 + 1 2 = 4 2 + (4-y) 2
⇒ 25 + 9 - 6y + y 2 + 1 + 1 = 16 + 16 - 8y + y 2
⇒ 36 + 2y - 32 = 0
⇒ 2y + 4 = 0
⇒ y = -2
16. Show that the points (a,a), (-a, -a) and (-√3a, √3a) are the vertices of an equilateral triangle.
Let A(a,a), B(-a, -a) and C(√3a, √3a)
Since AB = BC = AC, therefore ABC is an equilateral triangle.
17. If A(4, 3), B(-1, y) and C(3, 4) are the vertices of a right triangle ABC, right-angled at A, then find the value of y.
We have A(4, 3), B(-1, y) and C(3, 4). In right angled triangle ABC, (BC) 2 = (AB) 2 + (AC) …(Pythagoras theorem) ⇒ (-1 – 3) 2 + (y – 4) 2 = (4 + 1) 2 + (3 – y) 2 + (4 – 3) 2 + (3 – 4) 2 …(using distance formula) ⇒ (-4) 2 + (y 2 – 8y + 16) ⇒ (5) 2 + (9 – 6y + y 2 ) + (1) 2 + (-1) 2 ⇒ y 2 – 8y + 32 = y 2 – 6y + 36 = 0 ⇒ -8y + 6y + 32 – 36 ⇒ -2y – 4 = 0 ⇒ -2y = 4 ∴ y = -2
18. If the mid-point of the line segment joining A(x/2, y+1/2) and B(x+1, y-3) is C(5, -2), find x,y.
If the mid-point of the line segment joining A(x/2, y+1/2) and B(x+1, y-3) is C(5, -2) then at mid point,
⇒ y + 1 + 2y - 6 = -8
⇒ y = -1
19. Show that A(6,4) B(5,-2) and C(7,-2) are the vertices of an isosceles triangle.
We have A(6,4) B(5,-2) and C(7,-2).
Since two sides of a triangle are equal in length, triangle is an isosceles triangle.
20. If the line segment joining the points A(2,1) and B(5,-8) is trisected at the points P and Q, find the coordinates P.
As per question, line diagram is shown below.
Let P(x, y) divides AB in the ratio 1:2.
Using section formula we get,
Hence, coordinates of P are (3,-2).
21. Three vertices of a parallelogram taken in order are (-1, 0), (3, 1) and (2, 2) respectively. Find the coordinates of fourth vertex. (2011D)
Let A(-1, 0), B(3, 1), C(2, 2) and D(x, y) be the vertices of a parallelogram ABCD taken in order. Since, the diagonals of a parallelogram bisect each other. ∴ Coordinates of the mid-point of AC = Coordinates of the mid-point of BD
Hence, coordinates of the fourth vertex, D(-2, 1).
22. If two adjacent vertices of a parallelogram are (3,2) and (-1,0) and the diagonals intersect at (2,-5) then find the co-ordinates of the other two vertices.
Let two other co-ordinates be (x,y) and (x',y') respectively using mid-point formula.
As per question parallelogram is shown below.
Hence, coordinates of C(1,-12) and D(5,-10).
23. If A(5, 2), B(2, -2) and C(-2, t) are the vertices of a right angled triangle with ∠B = 90°, then find the value of t. (2015D)
ABC is a right angled triangle, ∴ AC 2 = BC 2 + AB 2 …(i) …(Pythagoras theorem) Using distance formula, AB 2 = (5 – 2) 2 + (2 + 2) = 25 BC 2 = (2 + 2) 2 + (t + 2) 2 = 16 + (t + 2) 2 AC 2 = (5 + 2) 2 + (2 – t) 2 = 49 + (2 – t) 2 Putting values of AB 2 , AC 2 and BC 2 in equation (i), we get 49 + (2 – t) 2 = 16 + (t + 2) 2 + 25 ∴ 49 + (2 – t) 2 = 41 + (t + 2) 2 ⇒ (t + 2) 2 – (2 – t) 2 = 8 ⇒ (t 2 + 4 + 4t – 4 – t 2 + 4t) = 8 8t = 8 ⇒ t = 1
24. If P(2,-1), Q(3,4), R(-2,3) and S(-3,-2) be four points in a plane, show that PQRS is a rhombus but not a square.
P(2,-1), Q(3,4), R(-2,3) and S(-3,-2)
Since, all the four sides are equal, PQRS is a rhombus.
Since, PQR is not a right triangle, PQRS is a rhombus but not a square.
Coordinate Geometry Class 10 Maths Important Questions Short Answer-II (3 Marks)
25. Find the ratio in which the segment joining the points (1,-3) and (4,5) is divided by x-axis? Also find the coordinates of this point on x-axis.
Let the required ratio be k:1 and the point on x-axis be (x,0).
(x 1 , y 1 ) = (1, -3)
and (x 2 , y 2 ) = (4,5)
Using section formula y coordinate we obtain,
Hence, the required ratio is 3/5 i.e. 3:5.
Now, again using section formula for x, we obtain
Coordinate of P is (117/8, 0).
26. If the point C(-1, 2) divides internally the line segment joining A(2,5) and B(x, y) in the ratio 3 : 4 find the coordinates of B.
From the given information we have drawn the figure as below.
3y + 20 = 14
⇒ 3y = 14 - 20 = -6
Hence, the coordinates of B(x, y) is (-5,-2).
27. If the co-ordinates of points A and B are (-2, -2) and (2, -4) respectively, find the co-ordinates of P such that AP = 3/7 AB, where P lies on the line segment AB.
AP = 3/7 AB
⇒ AP:PB = 3:4
PQ = 10 …(Given) ⇒ PQ 2 = 10 2 = 100 …(Squaring both sides) (9 – x) 2 + (10 – 4) 2 = 100 …(using distance formula) ⇒ (9 – x) 2 + 36 = 100 ⇒ (9 – x) 2 = 100 – 36 = 64 ⇒ (9 – x) = ± 8 …(Taking square-root on both sides) ⇒ 9 – x = 8 or 9 – x = -8 ⇒ 9 – 8 = x or 9+ 8 = x ⇒ x = 1 or x = 17
29. Find the value of y for which the distance between the points A (3,-1) and B (11, y) is 10 units.
AB = 10 units …(Given) ⇒ AB 2 = 10 2 = 100 …(Squaring both sides) (11 – 3) 2 + (y + 1) 2 = 100 ⇒ 8 2 + (y + 1) 2 = 100 ⇒ (y + 1) 2 = 100 – 64 = 36 ⇒ y + 1 = ±6 …(Taking square-root on both sides) ⇒ y = -1 ± 6 ∴ y = -7 or 5
30. The co-ordinates of the vertices of △ABC are A(7, 2), B(9, 10) and C(1, 4). If E and F are the mid-points of AB and AC respectively, prove that EF = 1/2 BC.
Let the mid-points of AB and AC be E(x 1 , y 1 ) and F(x 2 , y 2 ). As per question, triangle is shown below.
32. Find the point on y-axis which is equidistant from the points (5, -2) and (-3, 2).
Let the point be (0,y).
5 2 + (y+2) 2 = 3 2 + (y-2) 2
or, y 2 + 25 + 4y + 4 = 9 - 4y + 4
8y = -16 or, y = -2
or, Point (0, -2)
33. The vertices of △ABC are A(6, -2), B(0, -6) and C(4,8). Find the co-ordinates of mid-points of AB, BC and AC.
Let mid-point of AB, BC and AC be D(x 1 , y 1 ), E(x 2 , y 2 ) and F(x 3 , y 3 ). As per question, triangle is shown below.
Using section formula, the co-ordinates of the points D, E, F are:
The co-ordinates of the mid-points of AB, BC and AC are D(3, -4), E(2,1) and F(5,3) respectively.
PA = QA …(Given) ⇒ PA 2 = QA 2 …(Squaring both sides) (3 – 6) 2 + (y – 5) 2 = (3 – 0) 2 + (y + 3) 2 ⇒ 9 + (y – 5) 2 = 9 + (y + 3) 2 ⇒ (y – 5) 2 = (y + 3) 2 ⇒ y – 5 = ±(y + 3) …(Taking square root of both sides) ⇒ y – 5 = y + 3 y – 5 = -y – 3 0 = 8 …(not possible) ∴ y = 1
35. If the point P(x, y) is equidistant from the points A(a + b, b – a) and B(a – b, a + b), prove that bx = ay.
PA = PB …(Given) PA 2 = PB 2 …(Squaring both sides) ⇒ [(a+b) – x] 2 + [(b-a) – y) 2 = [(a–b) – x] 2 + [(a+b) – y] 2 ⇒ (a+b) 2 + x 2 – 2(a+b)x + (b–a) 2 + y 2 – 2(b–a)y = (a–b) 2 + x 2 – 2(a–b)x + (a+b) 2 + y 2 – 2(a+b)y …[∵ (a–b) 2 = (b–a) 2 ] ⇒ -2(a + b)x + 2(a – b)x = -2(a + b)y + 2(b – a)y ⇒ 2x(-a – b + a – b) = 2y(-a – b + b – a) ⇒ -2bx = –2ay ⇒ bx = ay
36. If the point P(k – 1, 2) is equidistant from the points A(3, k) and B(k, 5), find the values of k.
PA = PB …(Given) PA 2 = PB 2 …(Squaring both sides) ⇒ (k – 1 – 3) 2 + (2 – k) 2 = (k – 1 – k) 2 + (2 – 5) 2 ⇒ (k – 4) 2 + (2 – k) 2 = (-1) 2 + (-3) 2 ⇒ k 2 – 8k + 16 + 4 + k 2 – 4k = 1 + 9 ⇒ 2k 2 – 12k + 20 – 10 = 0 ⇒ 2k 2 – 12k + 10 = 0 …(Dividing by 2) ⇒ k 2 – 6k + 5 = 0 ⇒ k 2 – 5k – k + 5 = 0 ⇒ k(k – 5) – 1(k – 5) = 0 ⇒ (k – 5) (k – 1) = 0 ⇒ k – 5 = 0 or k – 1 = 0 ∴ k = 5 or k = 1
37. Find the area of the rhombus of vertices (3,0), (4,5), (-1,4) and (-2,-1) taken in order.
A(3,0), B(4,5), C(-1,4) and D(-2,-1)
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- Chapter 14 Co Ordinate Geometry
RD Sharma Solutions for Class 10 Maths Chapter 14 Co-ordinate Geometry
Rd sharma solutions class 10 maths chapter 14 – free pdf download.
The RD Sharma Solutions for Class 10 Maths Chapter 14 – Co-ordinate Geometry explain each concept in a comprehensive manner to help students obtain a thorough knowledge of concepts. Coordinates are ordered pairs of numbers, and the study of Geometry using coordinates is called Co-ordinate Geometry. Class 10 Mathematics explains planes and graphical interpretations. The faculty at BYJU’S have created the RD Sharma Solutions to boost problem-solving skills among students.
Chapter 14 Co-ordinate Geometry consists of five exercises. RD Sharma Solutions for Class 10 Maths provide descriptive answers for the problems of these exercises. The main purpose of preparing these solutions is to boost confidence in solving any type of problem with ease among students. Students can access the solutions PDF updated for 2023-24 from the links given here. Important concepts discussed in this chapter are as follows:
- To find the distance between two points whose coordinates are given.
- Find the coordinates of the point which divides the line segment joining two given points in a given ratio.
- The method of finding the area of a triangle in terms of the coordinates of its vertices.
- RD Sharma Solutions Class 10 Maths Chapter 1 Real Numbers
- RD Sharma Solutions Class 10 Maths Chapter 2 Polynomials
- RD Sharma Solutions Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables
- RD Sharma Solutions Class 10 Maths Chapter 4 Triangles
- RD Sharma Solutions Class 10 Maths Chapter 5 Trigonometric Ratios
- RD Sharma Solutions Class 10 Maths Chapter 6 Trigonometric Identities
- RD Sharma Solutions Class 10 Maths Chapter 7 Statistics
- RD Sharma Solutions Class 10 Maths Chapter 8 Quadratic Equations
- RD Sharma Solutions Class 10 Maths Chapter 9 Arithmetic Progressions
- RD Sharma Solutions Class 10 Maths Chapter 10 Circles
- RD Sharma Solutions Class 10 Maths Chapter 11 Constructions
- RD Sharma Solutions Class 10 Maths Chapter 12 Some Applications of Trigonometry
- RD Sharma Solutions Class 10 Maths Chapter 13 Probability
- RD Sharma Solutions Class 10 Maths Chapter 14 Co-ordinate Geometry
- RD Sharma Solutions Class 10 Maths Chapter 15 Areas Related to Circles
- RD Sharma Solutions Class 10 Maths Chapter 16 Surface Areas and Volumes
- Exercise 14.1 Chapter 14 Co-ordinate Geometry
- Exercise 14.2 Chapter 14 Co-ordinate Geometry
- Exercise 14.3 Chapter 14 Co-ordinate Geometry
- Exercise 14.4 Chapter 14 Co-ordinate Geometry
- Exercise 14.5 Chapter 14 Co-ordinate Geometry
Download the PDF of RD Sharma Solutions for Class 10 Maths Chapter 14 Co-ordinate Geometry Here
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Access the RD Sharma Solutions for Class 10 Maths Chapter 14 – Co-ordinate Geometry
Rd sharma solutions for class 10 maths chapter 14 exercise 14.1 page no: 14.4.
1. On which axis do the following points lie?
(i) P (5, 0)
(ii) Q (0, -2)
(iii) R (-4, 0)
(iv) S (0, 5)
(i) P (5, 0) lies on the x-axis
(ii) Q (0, -2) lies on the y-axis (negation half)
(iii) R (-4, 0) lies on the x-axis (negative half)
(iv) S (0, 5) lies on the y-axis
RD Sharma Solutions for Class 10 Maths Chapter 14 Exercise 14.2 Page No: 14.15
1. Find the distance between the following pair of points.
(i) (- 6, 7) and (-1, -5)
(ii) (a + b, b + c) and (a -b, c – b)
(iii) (a sin α, – b cos α) and (- a cos α, b sin α)
(iv) (a, 0) and (0, b)
(i) Let the given points be P (- 6, 7) and Q (- 1, – 5).
x 1 = – 6, y 1 = 7 and
x 2 = -1, y 2 = – 5
(ii) Let the given points be P (a + b, b + c) and Q (a – b, c – b).
x 1 = a + b, y 1 = b + c and
x 2 = a – b, y 2 = c – b
(iii) Let the given points be P(a sinα, – b cos α) and Q(-a cos α, b sin α).
x 1 = a sin α, y 1 = – b cos α and
x 2 – a cos α, y 2 = b sin α
(iv) Let the given points be P(a, 0) and Q (0, b).
x 1 = a, y 1 = 0, x 2 = 0, y 2 = b,
2. Find the value of a when the distance between the points (3, a) and (4, 1) is √10.
Let the given points be P (3, a) and Q(4, 1).
On squaring on both sides, we have
⇒ 10 = 2 + a 2 – 2a
⇒ a 2 – 2a + 2 – 10 = 0
⇒ a 2 – 2a – 8 = 0
By splitting the middle team,
⇒ a 2 – 4a + 2a – 8 = 0
⇒ a(a – 4) + 2(a – 4) = 0
⇒ (a – 4) (a + 2) = 0
⇒ a = 4, a = – 2
Thus, there are 2 possible values for a which are 4 and -2.
3. If the points (2, 1) and (1, -2) are equidistant from the point (x, y), show that x + 3y = 0.
Let the given points be P(2, 1) and Q(1,- 2) and R(x, y).
Also, PR = QR (given)
But, PR = QR
⇒ x 2 + 5 – 4x + y 2 –2y = x 2 + 5 – 2x + y 2 + 4y
⇒ x 2 + 5 – 4x + y 2 – 2y = x 2 + 5 – 2x + y 2 + 4y
⇒ – 4x + 2x – 2y – 4y = 0
⇒ – 2x – 6y = 0
⇒ – 2(x + 3y) = 0
⇒ -2(x + 3y) = 0
⇒ x + 3y = 0/-2
⇒ x + 3y = 0
Hence, proved.
4. Find the value of x, y if the distances of the point (x, y) from (- 3, 0) as well as from (3, 0) are 4.
Let the given points be P(x, y), Q( -3, 0) and R(3, 0).
On squaring on both sides, we get
⇒ 16 = x 2 + 9 + 6x + y 2
⇒ x 2 + y 2 = 7 – 6x …… (1)
On squaring on both sides,
⇒16 = x 2 + 9 – 6x + y 2
⇒ x 2 + y 2 = 16 – 9 + 6x
⇒ x 2 + y 2 = 7 + 6x …. (2)
Equating (1) and (2), we have
7 – 6x = 7 + 6x
⇒ 7 – 7 = 6x + 6x
⇒ 0 = 12x
⇒ x = 12
Then, substituting the value of x = 0 in (2),
x 2 + y 2 = 7+ 6x
0 + y 2 = 7 + 6 × 0
y 2 = 7
y = + √7
As y can have two values, the points are (12, √7) and (12, -√7).
5. The length of a line segment is of 10 units, and the coordinates of one end-point are (2, -3). If the abscissa of the other end is 10, find the ordinate of the other end.
The length of the line segment is 10 units.
The coordinates of one end-point are (2, -3), and the abscissa of the other end is 10.
So, let the ordinate of the other end be k.
Therefore, the ordinates of the other end can be 3 or -9.
6. Show that the points A(- 4, -1), B(-2, – 4), C(4, 0) and D(2, 3) are the vertices points of a rectangle.
Given: Points A(- 4, -1), B(-2, – 4), C(4, 0) and D(2, 3)
Required to prove: the points are the vertices points of a rectangle.
Vertices of rectangle ABCD are A(- 4, -1), B(-2, – 4), C(4, 0) and D(2, 3)
We know that
As the opposite sides are equal and also the diagonals are equal.
Therefore, the given points are the vertices of a rectangle.
7. Show that the points A (1,- 2), B (3, 6), C (5, 10) and D (3, 2) are the vertices of a parallelogram.
Given: Points A (1,- 2), B (3, 6), C (5, 10) and D (3, 2)
Required to prove: the points are the vertices points of a parallelogram.
Vertices of a parallelogram ABCD are A (1, -2), B (3, 6), C (5, 10) and D (3, 2)
Finding the diagonals,
It’s seen that the opposite sides of the quadrilateral formed by the given four points are equal
i.e., (AB = CD) & (DA = BC)
Also, the diagonals BD and AC are found to be unequal.
Hence, the given points form a parallelogram.
8. Prove that the points A (1, 7), B (4, 2), C (-1, -1) and D (-4, 4) are the vertices of a square.
Given: Points A (1, 7), B (4, 2), C (-1, -1) and D (-4, 4)
Required to prove: the points are the vertices points of a square.
Vertices of a square ABCD are A (1, 7), B (4, 2), C (-1, -1) and D (-4, 4)
As the opposite sides are equal and also the diagonals are equal, the given vertices are, therefore, the vertices of a square.
9. Prove that the points (3, 0), (6, 4) and (- 1, 3) are vertices of a right-angled isosceles triangle.
Let the vertices of the triangle ABC be A(3, 0), B(6, 4) and C (- 1, 3)
We know that,
It’s seen that AB = AC. Thus, it’s an isosceles triangle.
Verifying the Pythagoras theorem, we have
BC 2 = AB 2 + AC 2
(√50) 2 = (√25) 2 + (√25) 2
50 = 25 + 25
As BC 2 = AB 2 + AC 2
Therefore, the given vertices are of a right-angled isosceles triangle.
10. Prove that (2, -2), (-2, 1) and (5, 2) are the vertices of a right-angled triangle. Find the area of the triangle and the length of the hypotenuse.
From the given question,
Let consider the vertices of a triangle ABC as A(2, -2), B(-2, 1) and C(5, 2)
It’s seen that AB = AC, thus, the triangle is an isosceles triangle.
Therefore, the given triangle is right angled triangle.
11. Prove that the points (2a, 4a), (2a, 6a) and (2a + √3a, 5a ) are the vertices of an equilateral triangle.
From given,
Let’s consider the vertices of a triangle ABC as A(2 a, 4 a), B(2 a, 6 a) and C(2a + √3a, 5a)
As all the sides are equal, the triangle is an equilateral triangle.
Thus, the given vertices are of an equilateral triangle.
12. Prove that the points (2, 3), (-4, -6) and (1, 3/2) do not form a triangle.
Let’s consider the vertices of a triangle ABC as A(2, 3), B(-4, -6) and C(1, 3/2)
Thus, the given vertices do not form a triangle as the sum of two sides of a triangle is not greater than the third side.
13. The points A (2, 9), B (a, 5) and C (5, 5) are the vertices of a triangle ABC, right-angled at B. Find the values of a and hence the area of triangle ABC.
A right triangle ABC, right-angled at B.
Points A (2, 9), B (a, 5) and C (5, 5)
So, AC is the hypotenuse.
Thus, from Pythagoras’ theorem, we have
AC 2 = AB 2 + BC 2
9 + 16 = a 2 – 4a + 4 + 16 + 25 – 10a + a 2
2a 2 – 14a + 20 = 0
a 2 – 7a + 10 = 0
(a – 5)(a – 2) = 0 [By factorisation method]
Here, a = 5 is not possible as it coincides with point C. So, for a triangle to form, the value of a = 2 is correct.
Thus, the coordinates of point B are (2, 5).
Now, the area of triangle ABC.
Therefore, the area of triangle ABC is 6 sq. units.
14. Show that the quadrilateral whose vertices are (2, -1), (3, 4), (-2, 3) and (-3, -2) is a rhombus.
Let A(2, -1), B(3 ,4), C(-2, 3) and D(-3, -2)
Then, we have,
Length of AB = √[(3 – 2) 2 + (4 – (-1)) 2 ] = √[(1) 2 + (5) 2 ] = √[1 + 25] = √26 units
Length of BC = √[(3 – (-2)) 2 + (4 – 3) 2 ] = √[(5) 2 + (1) 2 ] = √[25 + 1] = √26 units
Length of CD = √[(-2 – (-3)) 2 + (3 – 2) 2 ] = √[(-5) 2 + (1) 2 ] = √[25 + 1] = √26 units
Length of AD = √[(-3 – 2) 2 + (-2 – (-1)) 2 ] = √[(-5) 2 + (-1) 2 ] = √[25 + 1] = √26 units
As AB = BC = CD = AD
We can say that
Quadrilateral ABCD is a rhombus.
15. Two vertices of an isosceles triangle are (2, 0) and (2, 5). Find the third vertex if the length of the equal sides is 3.
Let the third vertex be C (x, y).
And given A (2, 0) and B (2, 5).
Length of AB = √[(2 – 2) 2 + (5 – 0) 2 ] = √[(0) 2 + (5) 2 ] = √[0 + 25] = 5 units
Length of BC = √[(x – 2) 2 + (y – 5) 2 ] = √[x 2 – 4x + 4 + y 2 – 10y + 25]
= √[ x 2 – 4x + y 2 – 10y + 29] units
Length of AC = √[(x – 2) 2 + (y – 0) 2 ] = √[x 2 – 4x + 4 + y 2 ] units
AC = BC = 3
So, AC 2 = BC 2 = 9
x 2 – 4x + 4 + y 2 = x 2 – 4x + y 2 – 10y + 29
y = 25/10 = 2.5
x 2 – 4x + 4 + y 2 = 9
x 2 – 4x + 4 + (2.5) 2 = 9
x 2 – 4x + 4 + 6.25 = 9
x 2 – 4x + 1.25 = 0
D = (-4) 2 – 4 x 1 x 1.25 = 16 – 5 = 11
So, the roots are
x = -(-4) + √11/ 2 = (4 + 3.31)/ 2 = 3.65
x = -(-4) – √11/ 2 = (4 – 3.31)/ 2 = 0.35
Therefore, the third vertex can be C (3.65, 2.5) or (0.35, 2.5).
16. Which point on the x-axis is equidistant from (5, 9) and (-4, 6)?
Let A (5, 9) and B (-4, 6) be the given points.
Let the point on the x-axis equidistant from the above points be C(x, 0).
Now, we have
AC = √[(x – 5) 2 + (0 – 9) 2 ] = √[x 2 – 10x + 25 + 81] = √[x 2 – 10x + 106]
BC = √[(x – (-4)) 2 + (0 – 6) 2 ] = √[x 2 + 8x + 16 + 36] = √[x 2 + 8x + 52]
As AC = BC (given condition)
So, AC 2 = BC 2
x 2 – 10x + 106 = x 2 + 8x + 52
Therefore, the point on the x-axis is (3, 0).
17. Prove that the points (-2, 5), (0, 1) and (2, -3) are collinear.
Let A (-2, 5), B(0, 1) and C (2, -3) be the given points.
So, we have
AB = √[(0 – (-2)) 2 + (1 – 5) 2 ] = √[(2) 2 + (-4) 2 ] = √[4 + 16] = √20 = 2√5 units
BC = √[(2 – 0) 2 + (-3 – 1) 2 ] = √[(2) 2 + (-4) 2 ] = √[4 + 16] = √20 = 2√5 units
AC = √[(2 – (-2)) 2 + (-3 – 5) 2 ] = √[(4) 2 + (-8) 2 ] = √[16 + 64] = √80 = 4√5 units
Now, it’s seen that
AB + BC = AC
2√5 + 2√5 = 4√5
4√5 = 4√5
Therefore, we can conclude that the given points (-2, 5), (0, 1) and (2, -3) are collinear.
18. The coordinates of point P are (-3, 2). Find the coordinates of the point Q, which lies on the line joining P and origin such that OP = OQ.
Let the coordinates of Q be taken as (x, y).
As Q lies on the line joining P and O(origin) with OP = OQ.
Then, by mid-point theorem,
(x – 3)/2 = 0
(y + 2)/ 2 = 0
∴ x = 3, y = -2
Therefore, the coordinates of point Q are (3, -2).
19. Which point on the y-axis is equidistant from (2, 3) and (-4, 1)?
Let A (2, 3) and B (-4, 1) be the given points.
Let the point on the y-axis equidistant from the above points be C (0, y).
AC = √[(0 – 2) 2 + (y – 3) 2 ] = √[y 2 – 6y + 9 + 4] = √[y 2 – 6y + 13]
BC = √[(0 – (-4)) 2 + (y – 1) 2 ] = √y 2 – 2y + 1 + 16] = √[y 2 – 2y + 17]
y 2 – 6y + 13 = y 2 – 2y + 17
Therefore, the point on the y-axis is (0, -1).
20. The three vertices of a parallelogram are (3, 4), (3, 8) and (9, 8). Find the fourth vertex.
Let A (3, 4), B (3, 8) and C (9, 8) be the given points.
And let the fourth vertex be D(x, y).
In a parallelogram, the diagonals bisect each other.
So, the mid-point of AC should be the same as the mid-point of BC.
By mid-point theorem,
Mid-point of AC = (3 + 9/ 2), (4 + 8/ 2) = (6, 6)
The mid-point of BD = (3 + x/ 2, 8 + y/ 2)
And this point must be equal to (6, 6).
(3 + x)/ 2 = 6 (8 + y)/ 2 = 6
3 + x = 12 8 + y = 12
x = 9 y = 4
Therefore, the fourth vertex is D (9, 4).
21. Find a point which is equidistant from points A (-5, 4) and B (-1, 6). How many such points are there?
Let P(x, y) be the equidistant point from points A (-5, 4) and B (-1, 6).
So, the mid-point can be the required point.
(x, y) = ( (-5 – 1/ 2), (4 + 6)/2 )
(x , y) = (-6/2 , 10/2) = (-3, 5)
Thus, the required point is (-3, 5)
We also know that AP = BP
So, AP 2 = BP 2
(x + 5) 2 + (y – 4) 2 = (x + 1) 2 + (y – 6) 2
x 2 + 25 + 10 + y 2 – 8y + 16 = x 2 + 2x + 1 + y 2 – 12y + 36
10x + 41 – 8y = 2x + 37 – 12y
8x + 4y + 4 = 0
2x + y + 1 = 0
Therefore, all the points which lie on line 2x + y + 1 = 0 are equidistant from A and B.
22. The centre of a circle is (2a, a – 7). Find the values of a if the circle passes through the point (11, -9) and has diameter 10√2 units.
Diameter of the circle = 10√2 units
So, the radius = 5√2 units
Let the centre of a circle be 0(2a, a-7), and the circle passes through point P (11, -9).
Then, OP is the radius of the circle.
OP = 5√2
OP 2 = (5√2) = 50
(11- 2a) 2 + (-9 – a + 7) 2 = 50
121 – 44a + 4a 2 + 4 + a 2 + 4a = 50
5a 2 – 40a + 75 = 0
a 2 – 8a + 15 = 0
(a – 5)(a – 3) = 0 [Factorisation method]
So, a = 5 or a = 3
23. Ayush starts walking from his house to the office. Instead of going to the office directly, he goes to the bank first, from there to his daughter’s school and then reaches the office. What is the extra distance travelled by Ayush in reaching the office? (Assume that all distances covered are in straight lines). If the house is situated at (2, 4), bank at (5, 8), school at (13, 14) and office at (13, 26) and coordinates are in kilometers.
The position of Ayush’s house is (2, 4), and the position of the bank is (5, 8).
So, the distance between the house and the bank,
d 1 = √[(5 – 2) 2 + (8 – 4) 2 ] = √[(3) 2 + (4) 2 ] = √[9 + 16] = √25 = 5 km
The position of the bank is (5, 8), and the position of the school is (13, 14).
So, the distance between the bank and the school
d 2 = √[(13 – 5) 2 + (14 – 8) 2 ] = √[(8) 2 + (6) 2 ] = √[64 + 36] = √100 = 10 km
The position of the school is (13, 14), and the position of the office is (13, 26).
So, the distance between the school and the office is
d 3 = √[(13 – 13) 2 + (26 – 14) 2 ] = √[(0) 2 + (12) 2 ] = √144 = 12 km
Let d be the total distance covered by Ayush.
d = d 1 + d 2 + d 3 = 5 + 10 + 12 = 27 km
Let the D be the shortest distance from Ayush’s house to the office.
D = √[(13 – 2) 2 + (26 – 4) 2 ] = √[(11) 2 + (22) 2 ] = √[121+ 484] = √605 = 24.6 km
Thus, the extra distance covered by Ayush = d – D = 27 – 24.6 = 2.4 km.
24. Find the value of k, if the point P(0, 2) is equidistant from (3, k) and (k, 5).
Let the point P (0, 2) be equidistant from A (3, k) and B (k, 5).
So, PA = PB
PA 2 = PB 2
(3 -0) 2 + (k -2) 2 = (k – 0) 2 + (5 – 2) 2
9 + k 2 + 4 – 4k – k 2 – 9 = 0
4 – 4k = 0
Therefore, the value of k = 1.
25. If (-4, 3) and (4, 3) are two vertices of an equilateral triangle, find the coordinates of the third vertex, given that the origin lies in the (i) interior (ii) exterior of the triangle.
Let B (-4, 3) and C (4, 3) be the given two vertices of the equilateral triangle.
Let A (x, y) be the third vertex.
Then, we have
AB = BC = AC
Let us consider the part AB = BC.
AB 2 = BC 2
(-4 – x) 2 + (3 – y) 2 = (4 + 4) 2 + (3 – 3) 2
16 + x 2 + 8x + 9 + y 2 – 6y = 64
x 2 + y 2 + 8x – 6y = 39
Now, let us consider AB = AC.
AB 2 = AC 2
(-4 – x) 2 + (3 – y) 2 = (4 – x) 2 + (3 – y) 2
16 + x 2 + 8x + 9 + y 2 – 18y = 16 + x 2 – 8x + 9 + y 2 – 6y
Now, BC = AC
BC 2 = AC 2
(4 + 4) 2 + (3 – 3) 2 = (4 – 0) 2 + (3 – y) 2
64 + 0 = 16 + 9 + y 2 – 6y
64 = 16 + (3 – y) 2
(3 – y) 2 = 48
3 – y = ± 4√3
y = 3 ± 4√3
Therefore, the coordinates of the third vertex.
(i) When the origin lies in the interior of the triangle is (0, 3 – 4√3).
(ii) When the origin lies in the exterior of the triangle is (0, 3 + 4√3).
26. Show that the points (-3, 2), (-5, -5), (2, -3) and (4, 4) are the vertices of a rhombus. Find the area of this rhombus.
Let A(-3, 2), B(-5, -5), C(2, -3) and D(4, 4) be the given points.
AB = √[(-5 + 3) 2 + (-5 – 2) 2 ] = √[(2) 2 + (7) 2 ] = √[4 + 49] = √53 units
BC = √[(2 + 5) 2 + (-3 + 5) 2 ] = √[(7) 2 + (2) 2 ] = √[49 + 4] = √53 units
CD = √[(4 – 2) 2 + (4 + 3) 2 ] = √[(2) 2 + (7) 2 ] = √[4 + 49] = √53 units
AD = √[(4 + 3) 2 + (4 – 2) 2 ] = √[(7) 2 + (2) 2 ] = √[49 + 4] = √53 units
And the diagonals
AC = √[(2 + 3) 2 + (-3 – 2) 2 ] = √[(5) 2 + (-5) 2 ] = √[25 + 25] = 5√2 units
BD = √[(4 + 5) 2 + (4 + 5) 2 ] = √[(9) 2 + (9) 2 ] = √[81 + 81] = 9√2 units
It’s seen that
As AB = BC = CD = AD and the diagonals AC ≠ BD
ABCD is a rhombus.
Area of rhombus ABCD = ½ x AC x BD = ½ x 5√2 x 9√2 = 45 sq. units
27. Find the coordinates of the circumcenter of the triangle whose vertices are (3, 0), (-1, -6) and (4, -1). Also, find the circumradius.
Let A(3, 0), B(-1, -6) and C(4 , -1) be the given points.
Let O(x, y) be the circumcenter of the triangle.
Then, OA = OB = OC
OA 2 = OB 2
(x – 3) 2 + (y – 0) 2 = (x + 1) 2 + (y + 6) 2
x 2 + 9 – 6x + y 2 = x 2 + 1 + 2x + y 2 + 36 + 12y
-8x -12y = 28
2x + 3y = -7 …..(i) [After simplification]
OB 2 = OC 2
(x + 1) 2 + (y + 6) 2 = (x – 4) 2 + (y + 1) 2
x 2 + 2x + 1 + y 2 + 36 + 12y = x 2 + 16 – 8x + y 2 + 1 + 2y
10x + 10y = -20
x + y = -2 …..(ii) [After simplification]
Hence, the circumcenter of the triangle is (1, -3)
Circumradius = Distance from any of the given points (say B)
=√[(1 + 1) 2 + (-3 + 6) 2 ] = √(4 + 9)
= √13 units
28. Find a point on the x-axis which is equidistant from the points (7, 6) and (-3, 4).
Let A(7, 6) and B(-3, 4) be the given points.
Let P(x, 0) be the point on the x-axis such that PA = PB
So, PA 2 = PB 2
(x – 7) 2 + (0 – 6) 2 = (x + 3) 2 + (0 – 4) 2
x 2 + 49 – 14x + 36 = x 2 + 9 + 6x + 16
RD Sharma Solutions for Class 10 Maths Chapter 14 Exercise 14.3 Page No: 14.28
1. Find the coordinates of the point which divides the line segment joining (-1, 3) and (4, – 7) internally in the ratio 3: 4.
Let P(x, y) be the required point.
By section formula, we know that the coordinates are
x 1 = – 1 y 1 = 3
x 2 = 4 y 2 = -7
m: n = 3: 4
Therefore, the coordinates of P are (8/7, – 9/7).
2. Find the points of trisection of the line segment joining the points.
(i) (5, – 6) and (-7, 5)
(ii) (3, – 2) and (-3, – 4)
(iii) (2, – 2) and (-7, 4)
(i) Let P and Q be the point of trisection of AB, such that AP = PQ = QB.
So, P divides AB internally in the ratio of 1: 2, thereby applying the section formula, the coordinates of P will be
Now, Q also divides AB internally in the ratio of 2:1, so their coordinates will be
(ii) Let P and Q be the points of trisection of AB such that AP = PQ = QB
P divides AB internally in the ratio of 1: 2. Hence, by applying the section formula, the coordinates of P are
Now, Q also divides internally in the ratio of 2: 1
So, the coordinates of Q are given by
(iii) Let P and Q be the points of trisection of AB such that AP = PQ = OQ
P divides AB internally in the ratio 1:2. So, the coordinates of P, by applying the section formula, are given by
Now, Q also divides AB internally in the ration 2: 1. And the coordinates of Q are given by
3. Find the coordinates of the point where the diagonals of the parallelogram formed by joining the points (-2, -1), (1, 0), (4, 3) and (1, 2) meet.
Let A(-2, -1), B(1, 0), C(4, 3) and D(1, 2) be the given points.
Let P(x, y) be the point of intersection of the diagonals of the parallelogram formed by the given points.
We know that diagonals of a parallelogram bisect each other.
Therefore, the coordinates of P are (1, 1).
4. Prove that the points (3, 2), (4, 0), (6, -3) and (5, -5) are the vertices of a parallelogram.
Let A(3, -2), B(4, 0), C(6, -3) and D(5, -5).
Let P(x, y) be the point of intersection of diagonals AC and BD of ABCD.
The mid-point of AC is given by
Again, the mid-point of BD is given by
Thus, we can conclude that diagonals AC and BD bisect each other.
And we know that diagonals of a parallelogram bisect each other.
Therefore, ABCD is a parallelogram.
5. If P(9a – 2, -b) divides the line segment joining A(3a + 1, -3) and B(8a, 5) in the ratio 3 : 1, find the values of a and b.
Given that, P(9a – 2, -b) divides the line segment joining A(3a + 1, -3) and B(8a, 5) in the ratio 3:1
Then, by THE section formula
Coordinates of P are
Solving for a, we have
(9a – 2) x 4 = 24a + 3a + 1
36a – 8 = 27a + 1
Now, solving for b, we have
4 x –b = 15 – 3
Therefore, the values of a and b are 1 and -3, respectively.
6. If (a, b) is the mid-point of the line segment joining the points A (10, -6), B(k, 4) and a – 2b = 18, find the value of k and the distance AB.
As (a, b) is the mid-point of the line segment A(10, -6) and B(k, 4).
(a, b) = (10 + k / 2, -6 + 4/ 2)
a = (10 + k)/ 2 and b = -1
2a = 10 + k
k = 2a – 10
Given a – 2b = 18
Using b = -1 in the above relation, we get
a – 2(-1) = 18
a = 18 – 2 = 16
k = 2(16) – 10 = 32 – 10 = 22
AB = √[(22 – 10) 2 + (4 + 6) 2 ] = √[(12) 2 + (10) 2 ] = √[144 + 100] = 2√61 units
7. Find the ratio in which the point (2, y) divides the line segment joining the points A(-2, 2) and B(3, 7). Also, find the value of y.
Let the point P(2, y) divide the line segment joining the points A(-2, 2) and B(3, 7) in the ratio k: 1
Then, the coordinates of P are given by
And, given the coordinates of P are (2, y).
2 = (3k – 2)/ (k + 1) and y = (7k + 2)/ (k + 1)
Solving for k, we get
2(k + 1) = (3k – 2)
2k + 2 = 3k – 2
Using k to find y, we have
y = (7(4) + 2)/ (4 + 1)
= (28 + 2)/5
Therefore, the ratio is 4: 1 and y = 6
8. If A(-1, 3), B(1, -1) and C(5, 1) are the vertices of a triangle ABC, find the length of the median through A.
Let AD be the median through A.
AD is the median, AND D is the mid-point of BC.
So, the coordinates of D are (1 + 5/ 2, -1 + 1/ 2 ) = (3 , 0)
Length of median AD = √[(3 + 1) 2 + (0 – 3) 2 ] = √[(4) 2 + (-3) 2 ] = √[16 + 9] = √25 = 5 units
9. If the points P, Q(x, 7), R, S(6, y) in this order divide the line segment joining A(2, p) and B (7, 10) in 5 equal parts, find x, y and p.
From the question, we have
AP = PQ = QR = RS = SB
So, Q is the mid-point of A and S.
x = (2 + 6)/ 2 = 8/2 = 4
7 = (y + p)/ 2
y + p = 14 ….. (1)
Now, since S divides QB in the ratio 2: 1
So, p = 14 – 9 = 5
Therefore, x = 4, y = 9 and p = 5
10. If a vertex of a triangle is (1, 1) and the middle points of the sides through it be (-2, 3) and (5, 2), find the other vertices.
Let A(1, 1) be the given vertex and D(-2, 3) and E(5, 2) be the mid-points of AB and AC.
Now, D and E are the mid-points of AB and AC.
So, the coordinates of B are (-5, 5).
So, the coordinates of C are (9, 3).
Therefore, the other vertices of the triangle are (-5, 5) and (9, 3).
11. (i) In what ratio is the segment joining the points, (-2, -3) and (3, 7), divided by the y-axis? Also, find the coordinates of the point of division.
Let P(-2, -3) and Q(9, 3) be the given points.
Suppose the y-axis divides PQ in the ratio k: 1 at R(0, y).
So, the coordinates of R are given by
Now, equating,
3k – 2 = 0
Therefore, the ratio is 2: 3
Putting k = 2/3 in the coordinates of R, we get
(ii) In what ratio is the line segment joining (-3, -1) and (-8, -9) divided at the point (-5, -21/5)?
Let A(-3, -1) and B(-8, -9) be the given points.
And, let P be the point that divides AB in the ratio k: 1
So, the coordinates of P are given by
But, given the coordinates of P,
On equating, we get
(-8k – 3)/ (k + 1) = -5
-8k – 3 = -5k – 5
Thus, point P divides AB in the ratio 2: 3
12. If the mid-point of the line joining (3, 4) and (k, 7) is (x, y) and 2x + 2y + 1 = 0, find the value of k.
As (x, y) is the mid-point,
x = (3 + k)/ 2 and y = (4 + 7)/ 2 = 11/2
Given that the mid-point lies on line 2x + 2y + 1 = 0
2[(3 + k)/ 2] + 2(11/2) + 1 = 0
3 + k + 11 + 1 = 0
Thus, k = -15
13. Find the ratio in which the point P(3/4, 5/12) divides the line segments joining the point A(1/2, 3/2) and B(2, -5).
Points A(1/2, 3/2) and B(2, -5)
Let the point P(3/4, 5/12) divide the line segment AB in the ratio k: 1
Then, we know that
P(3/4, 5/12) = (2k + ½)/ (k +1) , (2k + 3/2)/ (k + 1)
Now, equating the abscissa, we get
¾ = (2k + ½)/ (k +1)
3(k + 1) = 4(2k + 1/2)
3k + 3 = 8k + 2
Therefore, the ratio in which the point P(3/4, 5/12) divides is 1: 5
14. Find the ratio in which the line joining (-2, -3) and (5, 6) is divided by (i) x-axis and (ii) y-axis. Also, find the coordinates of the point of division in each case.
Let A(-2, -3) and B(5, 6) be the given points.
(i) Suppose the x-axis divides AB in the ratio k: 1 at the point P.
Then, the coordinates of the point of division are
P lies on the x-axis, and the y-coordinate is zero.
6k – 3/ k + 1 = 0
6k – 3 = 0
k = ½
Thus, the required ratio is 1: 2
Using k in the coordinates of P,
We get P (1/3, 0)
(ii) Suppose the y-axis divides AB in the ratio k: 1 at point Q.
Then, the coordinates of the point od division are given by
Q lies on the y-axis, and the x-ordinate is zero.
5k – 2/ k + 1 = 0
5k – 2 = 0
Thus, the required ratio is 2: 5
Using k in the coordinates of Q,
We get Q (0, -3/7)
15. Prove that the points (4, 5), (7, 6), (6, 3), and (3, 2) are the vertices of a parallelogram. Is it a rectangle?
Let A (4, 5), B(7, 6), C(6, 3) and D(3, 2) be the given points.
And P be the point of intersection of AC and BD.
Coordinates of the mid-point of AC are (4+6/2 , 5+3/2) = (5, 4)
Coordinates of the mid-point of BD are (7+3/2 , 6+2/2) = (5, 4)
Thus, it’s clearly seen that the mid-point of AC and BD are the same.
So, ABCD is a parallelogram.
AC = √[(6 – 4) 2 + (3 – 5) 2 ] = √[(2) 2 + (-2) 2 ] = √[4 + 4] = √8 units
BD = √[(7 – 3) 2 + (6 – 2) 2 ] = √[(4) 2 + (4) 2 ] = √[16 + 16] = √32 units
Since AC ≠ BD
Therefore, ABCD is not a rectangle.
16. Prove that (4, 3), (6, 4), (5, 6) and (3, 5) are the angular points of a square.
Let A(4,3) , B(6,4) , C(5,6) and D(3,5) be the given points.
The distance formula is
It’s seen that the lengths of all the sides are the same.
Now, the length of diagonals are
Also, the lengths of both diagonals are the same.
Therefore, we can conclude that the given points are the angular points of a square.
17. Prove that the points (-4, -1), (-2, -4), (4, 0) and (2, 3) are the vertices of a rectangle.
Let A(-4, -1), B(-2, -4), C(4, 0) and D(2, 3) be the given points.
Coordinates of the mid-point of AC are (-4 + 4/ 2, -1 + 0/2) = (0, -1/2)
Coordinates of the mid-point of BD are (-2 + 2/2, -4 + 3/2) = (0, -1/2)
Thus, it’s seen that AC and BD have the same point.
And we have diagonals
The lengths of the diagonals are also the same.
18. Find the length of the medians of a triangle whose vertices are A(-1, 3), B(1, -1) and C(5, 1).
Let AD, BF and CE be the medians of ΔABC.
Coordinates of D are (5 + 1/ 2, 1 – 1/ 2) = (3, 0)
Coordinates of E are (-1 + 1/ 2, 3 – 1/ 2) = (0, 1)
Coordinates of F are (5 – 1/ 2, 1 + 3/ 2) = (2, 2)
Finding the length of the respective medians,
19. Find the ratio in which the line segment joining the points A (3, -3) and B (-2, 7) is divided by the x-axis. Also, find the coordinates of the point of division.
Let the point on the x-axis be (x, 0). [y-coordinate is zero].
And, let this point divides the line segment AB in the ratio of k : 1.
Now, using the section formula for the y-coordinate, we have
0 = (7k – 3)/(k + 1)
7k – 3 = 0
Therefore, the line segment AB is divided by the x-axis in the ratio 3: 7
20. Find the ratio in which point P(x, 2) divides the line segment joining the points A (12, 5) and B (4, -3). Also, find the value of x.
Let P divide the line joining A and B and let it divide the segment in the ratio k: 1
2 = (-3k + 5)/ (k + 1)
2(k + 1) = -3k + 5
2k + 2 = -3k + 5
Thus, P divides the line segment AB in the ratio of 3: 5
Using the value of k, we get the x-coordinate as
x = 12 + 60/ 8 = 72/8 = 9
Therefore, the coordinates of point P are (9, 2).
21. Find the ratio in which the point P(-1, y) lying on the line segment joining A(-3, 10) and B(6, -8) divides it. Also, find the value of y.
Let P divide A(-3, 10) and B(6, -8) in the ratio of k: 1.
Given coordinates of P as (-1, y).
Now, using the section formula for the x-coordinate, we have
-1 = 6k – 3/ k + 1
-(k + 1) = 6k – 3
Thus, point P divides AB in the ratio of 2: 7.
Using the value of k to find the y-coordinate, we have
y = (-8k + 10)/ (k + 1)
y = (-8(2/7) + 10)/ (2/7 + 1)
y = -16 + 70/ 2 + 7 = 54/9
Therefore, the y-coordinate of P is 6.
22. Find the coordinates of point A, where AB is the diameter of the circle whose centre is (2, -3) and B is (1, 4).
Let the coordinates of point A be (x, y).
If AB is the diameter, then the centre in the mid-point of the diameter,
(2, -3) = (x + 1/ 2, y + 4/ 2)
2 = x + 1/2 and -3 = y + 4/ 2
4 = x + 1 and -6 = y + 4
x = 3 and y = -10
Therefore, the coordinates of A are (3, -10).
23. If the points (-2, 1), (1, 0), (x, 3) and (1, y) form a parallelogram, find the values of x and y.
Let A(-2, 1), B(1, 0), C(x , 3) and D(1, y) be the given points of the parallelogram.
We know that the diagonals of a parallelogram bisect each other.
So, the coordinates of the mid-point of AC = Coordinates of the mid-point of BD
Therefore, the value of x is 4, and the value of y is 2.
24. The points A(2, 0), B(9, 1), C(11, 6) and D(4, 4) are the vertices of a quadrilateral ABCD. Determine whether ABCD is a rhombus or not.
Given points are A(2, 0), B(9, 1), C(11, 6) and D(4, 4).
Coordinates of mid-point of AC are (11+2/ 2, 6+0/ 2) = (13/2, 3)
Coordinates of mid-point of BD are (9+4/ 2, 1+4/ 2) = (13/2, 5/2)
As the coordinates of the mid-point of AC ≠ coordinates of the mid-point of BD, ABCD is not even a parallelogram.
Therefore, ABCD cannot be a rhombus too.
25. In what ratio does the point (-4,6) divide the line segment joining the points A(-6,10) and B(3,-8)?
Let the point (-4, 6) divide the line segment AB in the ratio k: 1.
So, using the section formula, we have
The same can be checked for the y-coordinate also.
Therefore, the ratio in which the point (-4, 6) divides the line segment AB is 2: 7
26. Find the ratio in which the y-axis divides the line segment joining the points (5, -6) and (-1, -4). Also, find the coordinates of the point of division.
Let P(5, -6) and Q(-1, -4) be the given points.
Let the y-axis divide the line segment PQ in the ratio k: 1.
Then, by using the section formula for the x-coordinate (as it’s zero), we have
Thus, the ratio in which the y-axis divides the given 2 points is 5: 1.
Now, to find the coordinates of the point of division,
Putting k = 5, we get
Hence, the coordinates of the point of division are (0, -13/3).
27. Show that A(-3, 2), B(-5, 5), C(2, -3) and D(4, 4) are the vertices of a rhombus.
Given points are A(-3, 2), B(-5, 5), C(2, -3) and D(4, 4).
Coordinates of the mid-point of AC are (-3+2/ 2, 2-3/ 2) = (-1/2, -1/2).
Coordinates of mid-point of BD are (-5+4/ 2, -5+4/ 2) = (-1/2, -1/2)
Thus, the mid-point for both diagonals is the same. So, ABCD is a parallelogram.
Next, the sides
It’s seen that ABCD is a parallelogram with adjacent sides equal.
Therefore, ABCD is a rhombus.
28. Find the lengths of the medians of a ΔABC having vertices at A(0, -1), B(2, 1) and C(0, 3).
Let AD, BE and CF be the medians of ΔABC.
Coordinates of D are (2+0/ 2, 1+3/ 2) = (1, 2)
Coordinates of E are (0/2, 3-1/ 2) = (0, 1)
Coordinates of F are (2+0/ 2, 1-1/ 2) = (1, 0)
Now, the length of the medians
29. Find the lengths of the median of a ΔABC having vertices at A(5, 1), B(1, 5) and C(-3, -1).
Given vertices of ΔABC as A(5, 1), B(1, 5) and C(-3, -1).
Let AD, BE and CF be the medians.
Coordinates of D are (1-3/ 2, 5-1/ 2) = (-1, 2)
Coordinates of E are (5-3/ 2, 1-1/2) = (1, 0)
Coordinates of F are (5+1/ 2, 1+5/ 2) = (3, 3)
30. Find the coordinates of the point which divide the line segment joining the points (-4, 0) and (0, 6) in four equal parts.
Let A(-4, 0) and B(0, 6) be the given points
And let P, Q and R be the points which divide AB is four equal points.
Now, we know that AP: PB = 1: 3
Using the section formula, the coordinates of P are
And it’s seen that Q is the mid-point of AB
So, the coordinates of Q are
Finally, the ratio of AR: BR is 3: 1
Then by using the section formula, the coordinates of R are
RD Sharma Solutions for Class 10 Maths Chapter 14 Exercise 14.4 Page No: 14.37
1. Find the centroid of the triangle whose vertices are:
(i) (1, 4), ( -1, -1) and (3, -2) (ii) (-2, 3), (2, -1) and (4, 0)
We know that the coordinates of the centroid of a triangle whose vertices are
(x 1 , y 1 ), (x 2 , y 2 ), (x 3 , y 3 ) are
(i) So, the coordinates of the centroid of a triangle whose vertices are
(1, 4), (-1, -1) and (3, -2) are
Thus, the centroid of the triangle is (1, 1/3).
(ii) So, the coordinates of the centroid of a triangle whose vertices are
(-2, 3), (2, -1) and (4, 0).
Thus, the centroid of the triangle is (4/3, 2/3)
2. Two vertices of a triangle are (1, 2), (3, 5), and its centroid is at the origin. Find the coordinates of the third vertex.
Let the coordinates of the third vertex be (x, y).
Then, we know that the coordinates of the centroid of the triangle are
Given that the centroid for the triangle is at the origin (0, 0).
⇒ x + 4 = 0 ⇒ y + 7 = 0
⇒ x = -4 ⇒ y = -7
Therefore, the coordinates of the third vertex are (-4, -7).
3. Find the third vertex of a triangle, if two of its vertices are at (-3, 1) and (0, -2) and the centroid is at the origin.
⇒ x – 3 = 0 ⇒ y – 1 = 0
⇒ x = 3 ⇒ y = 1
Therefore, the coordinates of the third vertex are (3, 1).
4. A(3, 2) and B(-2, 1) are two vertices of a triangle ABC whose centroid G has the coordinates (5/3, -1/3). Find the coordinates of the third vertex C of the triangle.
Let the coordinates of the third vertex C be (x, y).
Given, A(3, 2) and B(-2, 1) are two vertices of a triangle ABC.
Given that the centroid for the triangle is ( 5/3, -1/3).
⇒ x + 1 = 5 ⇒ y + 3 = -1
⇒ x = 4 ⇒ y = -4
Therefore, the coordinates of the third vertex C are (4, -4)
5. If (-2, 3), (4, -3) and (4, 5) are the mid-points of the sides of a triangle, find the coordinates of its centroid.
Let A (x 1 , y 1 ), B (x 2 , y 2 ) and C (x 3 , y 3 ) be the vertices of triangle ABC.
Let D (-2, 3), E (4, -3) and F (4, 5) be the mid-points of sides BC, CA and AB, respectively.
As D is the mid-point of BC
Similarly, E and F are the mid-points of AC and AB.
From (1), (2) and (3), we have
Forms (1) and (4), we get
Thus, the coordinates of A are (10, -1)
From (2) and (4), we get
Thus, the coordinates of B are (-2, 11).
From (3) and (4), we get
Thus, the coordinates of C are (-2, -5).
Hence, the vertices of triangle ABC are A (10, -1), B (-2, 11) and C (-2, -5).
Therefore, the coordinates of the centroid of triangle ABC are
RD Sharma Solutions for Class 10 Maths Chapter 14 Exercise 14.5 Page No: 14.53
1. Find the area of a triangle whose vertices are
(i) (6, 3), (-3, 5) and (4, – 2)
(ii) [(at 1 2 , at 1 ),( at 2 2 , 2at 2 )( at 3 2 , 2at 3 )]
(iii) (a, c + a), (a, c) and (-a, c – a)
(i) Let A(6, 3), B(-3, 5) and C(4,-2) be the given points
We know that the area of a triangle is given by
1/2[x 1 (y 2 – y 3 ) + x 2 (y 3 – y 1 ) + x 3 (y 1 + y 2 )]
x 1 = 6, y 1 = 3, x 2 = -3, y 2 = 5, x 3 = 4, y 3 = -2
Area of ∆ABC = 1/2 [6(5+2)+(-3)(- 2 -3)+ 4(3 – 5)]
=1/2 [6 × 7- 3 × ( – 5) + 4( – 2)]
= 1/2[42 +15 – 8]
= 49/2 sq. units
(ii) Let A = (x 1 , y 1 ) = (at 1 2 , 2at 1 ), B = (x 2 ,y 2 ) = (at 2 2 , 2at 2 ), C= (x 3 , y 3 ) = (at 3 2 , 2at 2 ) be the given points.
The area of ∆ABC is given by
(iii) Let A = (x 1 ,y 1 ) = (a, c + a), B = (x 2 , y 2 ) = (a, c) and C = (x 3 , y 3 ) = (- a, c – a) be the given points
= 1/2[a ( – {c – a}) + a(c – a – (c + a)) +( – a)(c + a – a)]
= 1/2 [a(c – c + a) + a(c – a – c – a) – a(c + a – c)]
= 1/2[a × a + ax( – 2a) – a × a]
= 1/2[a 2 – 2a 2 – a 2 ]
= 1/2×(-2a) 2
= – a 2
2. Find the area of the quadrilaterals, the coordinates of whose vertices are
(i) (-3, 2), (5, 4), (7, -6) and (-5, – 4)
(ii) (1, 2), (6, 2), (5, 3) and (3, 4)
(iii) (-4, -2), (-3, -5), (3, -2), (2, 3)
Let A(-3, 2), B(5, 4), C(7,- 6) and D ( -5, – 4) be the given points.
= 1/2[-3(4 + 6) + 5(- 6 – 2) + 7(2 – 4)]
= 1/2[-3×1 + 5×(-8) + 7(-2)]
= 1/2[- 30 – 40 -14]
= – 42
As the area cannot be negative,
The area of ∆ADC = 42 square units
Now, the area of ∆ADC is given by
= 1/2[-3( – 6 + 4) + 7(- 4 – 2) + (- 5)(2 + 6)]
= 1/2[- 3( – 2) + 7(- 6) – 5 × 8]
= 1/2[6 – 42 – 40]
= 1/2 × – 76
= – 38
But, as the area cannot be negative,
The area of ∆ADC = 38 square units
Thus, the area of quadrilateral ABCD = Ar. of ABC+ Ar. of ADC
= (42 + 38)
= 80 sq. units
Let A(1, 2), B (6, 2), C (5, 3) and (3, 4) be the given points.
Firstly, the area of ∆ABC is given by
= 1/2[1(2 – 3) + 6(3 – 2) + 5(2 – 2)]
= 1/2[ -1 + 6 × (1) + 0]
= 1/2[ – 1 + 6]
= 1/2[1(3 – 4) + 5(4 – 2) + 3(2 – 3)]
= 1/2[-1 × 5 × 2 + 3(-1)]
= 1/2[-1 + 10 – 3]
Thus, the area of quadrilateral ABCD = Area of ABC + Area of ADC
Let A (- 4, 2), B( – 3, – 5), C (3,- 2) and D(2, 3) be the given points.
= 1/2|(- 4)(- 5 + 2) – 3(-2 + 2) + 3(- 2 + 5)|
= 1/2|(-4)(-3) – 3(0) + 3(3)|
= 21/2
Now, the area of ∆ACD is given by
= 1/2|( – 4)(3 + 2) + 2( – 2 + 2) + 3( – 2 – 3)|
= 1/2|- 4(5) + 2(0) + 3(- 5)|= (- 35)/2
But, as the area can’t be negative,
The area of ∆ADC = 35/2
Thus, the area of quadrilateral (ABCD) = ar(∆ABC) + ar(∆ADC)
= 21/2 + 35/2
= 28 sq. units
3. The four vertices of a quadrilateral are (1, 2), (-5, 6), (7, -4) and (k, -2), taken in order. If the area of the quadrilateral is zero, find the value of k.
Let A(1, 2), B(-5, 6), C(7, -4) and D(k, -2) be the given points
= 1/2|(1)(6 + 4) – 5(-4 + 2) + 7(2 – 6)|
= 1/2|10 + 30 – 28|
= ½ x 12
= 1/2|(1)(-4 + 2) + 7( – 2 – 2) + k(2 + 4)|
= 1/2|- 2 + 7x(-4) + k(6)|
= (- 30 + 6k)/2
= 3k – 15
= 6 + 3k – 15
= 3k – 9
But, the given area of the quadrilateral is O.
So, 3k – 9 = 0
k = 9/3 = 3
4. The vertices of ΔABC are (-2, 1), (5, 4) and (2, -3), respectively. Find the area of the triangle and the length of the altitude through A.
Let A(-2, 1), B(5, 4) and C(2, -3) be the vertices of ΔABC.
And let AD be the altitude through A.
The area of ΔABC is given by
= 1/2|(-2)(4 + 3) – 5(-3 – 1) + 2(1 – 4)|
= 1/2|-14 – 20 – 6|
= ½ x -40
But as the area cannot be negative,
The area of ΔABC = 20 sq. units
We know that the area of a triangle is
= ½ x Base x Altitude
20 = ½ x √58 x AD
AD = 40/ √58
Therefore, the altitude AD = 40/ √58
5. Show that the following sets of points are collinear.
(a) (2, 5), (4, 6) and (8, 8) (ii) (1, -1), (2, 1) and (4, 5)
Condition: For the 3 points to be collinear, the area of the triangle formed with the 3 points has to be zero.
(a) Let A(2, 5), B(4, 6) and C(8, 8) be the given points.
Then, the area of ΔABC is given by
Since the area (ΔABC) = 0, the given points (2, 5), (4, 6) and (8, 8) are collinear.
(b) Let A(1, -1), B(2, 1) and C(4, 5) be the given points
Since the area (ΔABC) = 0, the given points (1, -1), (2, 1) and (4, 5) are collinear.
6. Find the area of a quadrilateral ABCD, the coordinates of whose vertices are A (-3, 2), B (5, 4), C (7, 6) and D (-5, -4).
Let’s join AC. So, we have 2 triangles formed.
Now, the ar (ABCD) = Ar (ΔABC) + Ar (ΔACD)
Next, the area of ΔACD is given by
Thus, the area (ABCD) = 42 + 38 = 80 sq. units
7. In ⧍ABC, the coordinates of vertex A are (0, -1) and D(1, 0) and E(0, 1), respectively, the mid-points of the sides AB and AC. If F is the mid-point of side BC, find the area of ⧍DEF.
Let B(a, b) and C(p, q) be the other two vertices of the ⧍ABC.
Now, we know that D is the mid-point of AB.
So, coordinates of D = (0+a/ 2, -1+b/ 2)
(1, 0) = (a/2, b-1/2)
1 = a/2 and 0 = (b-1)/ 2
a = 2 and b = 1
Hence, the coordinates of B = (2, 1)
E is the mid-point of AC.
So, coordinates of E = (0+p/ 2, -1+q/ 2)
(0, 1) = (p/2 , (q -1)/ 2)
p/2 = 0 and 1 = (q – 1)/2
p = 0 and 2 = q -1
p = 0 and q = 3
Hence, the coordinates of C = (0, 3)
Again, F is the mid-point of BC.
Coordinates of F = (2+0/ 2, 1+3/ 2) = (1, 2)
Thus, the area of ⧍DEF is given by
8. Find the area of the triangle PQR with Q (3, 2) and the mid-points of the sides through Q being (2, -1) and (1, 2).
Let the coordinates of P and R be (x 1 , y 1 ) and (x 2 , y 2 ), respectively.
And, let points E and F be the mid-points of PQ and QR, respectively.
x 1 + 3 = 2, y 1 + 2 = 4 and x 2 + 3 = 4, y 2 + 2 = -2
x 1 = -1, y 1 = 2 and x 2 = 1, y 2 = -4
Hence, the coordinates of P and R are (-1, 2) and (1, 0), respectively.
Therefore, the area of ⧍PQR is given by
9. If P(-5, -3), Q(-4, -6), R(2, -3) and S(1, 2) are the vertices of a quadrilateral PQRS, find its area.
First, let’s join P and R.
The area of ⧍PSR is given by
The area of ⧍PQR is given by
The area of the quad. PQRS = Area of ⧍PSR + Area of ⧍PQR
= 35/2 + 21/2
10. If A (-3, 5), B(-2, -7), C(1, -8) and D(6, 3) are the vertices of a quadrilateral ABCD, find its area.
Let’s join A and C.
So, we get ⧍ABC and ⧍ADC
The area of the quad. ABCD = Area of ⧍ABC + Area of ⧍ADC
Therefore, the area of the quadrilateral ABCD is 72 sq. units.
11. For what value of the points (a, 1), (1, -1) and (11, 4) are collinear?
Let A (a, 1), B (1, -1) and C (11, 4) be the given points.
Then the area of ⧍ABC is given by
We know that for the points to be collinear, the area of ⧍ABC has to be zero.
½(-5a + 25) = 0
∴ a = 5
12. Prove that the points (a, b), (a 1 , b 1 ) and (a-a 1 , b-b 1 ) are collinear if ab 1 = a 1 b
Let A (a, b), B (a 1 , b 1 ) and C (a-a 1 , b-b 1 ) be the given points.
So, the area of ⧍ABC is given by
So, only if ab 1 = a 1 b, the area becomes zero.
⧍ABC = ½ (0) = 0
Therefore, the given points are collinear if ab 1 = a 1 b
13. If the vertices of a triangle are (1,-3), (4,p) and (-9, 7) and its area is 15 sq. units, find the value (s) of p.
Let A(1,-3), B(4,p) and C(-9, 7) be the vertices of ⧍ABC.
Area of ⧍ABC = 15 sq. units
When the modulus is removed, two cases arise:
14. If (x, y) be on the line joining the two points (1, -3) and (-4, 2). Prove that x + y + 2 = 0
Let A (x, y), B (1, -3) and C (-4, 2) be the given points.
The area of ⧍ABC is given by,
The three points lie on the same line (that means they are collinear).
Then, the area of ⧍ABC = 0
½ (-5x – 5y – 10) = 0
-5x – 5y – 10 = 0
-5(x + y + 2) = 0
x + y + 2 = 0
15. Find the value of k if points (k, 3), (6, -2) and (-3, 4) are collinear.
Let A (k, 3), B (6, -2) and C (-3, 4) be the given points.
Then, the area of ⧍ABC is given by,
The points are collinear.
The area of ⧍ABC has to be zero.
½ x (-6k – 9) = 0
-6k – 9 = 0
∴ k = -3/2
16. Find the value of k, if points A(7, -2), B(5, 1) and C(3, 2k) are collinear.
Points A(7, -2), B(5, 1) and C(3, 2k)
½ (-4k + 8) = 0
-4k + 8 = 0
∴ k = 2
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Class 10 Maths Case Study Questions Chapter 7 Coordinate Geometry
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Case study Questions in the Class 10 Mathematics Chapter 7 are very important to solve for your exam. Class 10 Maths Chapter 7 Case Study Questions have been prepared for the latest exam pattern. You can check your knowledge by solving Class 10 Maths Case Study Questions Chapter 7 Coordinate Geometry
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In CBSE Class 10 Maths Paper, Students will have to answer some questions based on Assertion and Reason. There will be a few questions based on case studies and passage-based as well. In that, a paragraph will be given, and then the MCQ questions based on it will be asked.
Coordinate Geometry Case Study Questions With Answers
Here, we have provided case-based/passage-based questions for Class 10 Maths Chapter 7 Coordinate Geometry
Case Study/Passage-Based Questions
Question 1:
| (a) 12 cm | (b) 15 cm | (c) 18 cm | (d) none of these |
Answer: (d) none of these
(ii) The distance of the bus stand from the house is
| (a) 5 cm | (b) 10 cm | (c) 12 cm | (d) 15 cm |
Answer: (b) 10 cm
(iii) If the grocery store and electrician’s shop lie on a line, the ratio of the distance of the house from the grocery store to that from the electrician’s shop, is
| (a) 3.2 | (b) 2.3 | (c) 1.2 | (d) 2.1 |
Answer: (c) 1.2
(iv) The ratio of distances of the house from the bus stand to food cart is
| (a) 1.2 | (b) 2.1 | (c) 1.1 | (d) none of these |
Answer: (c) 1.1
(v) The coordinates of positions of bus stand, grocery store, food cart, and electrician’s shop form a
| (a) rectangle | (b) parallelogram | (c) square | (d) none of these |
Question 2:
The class X student’s school in krishnagar has been allotted a rectangular plot of land for their gardening activity. Saplings of Gulmohar are planted on the boundary at a distance of 1 m from each other. There is a triangular grassy lawn in the plot as shown in the figure. The students are to sow seeds of flowering plants on the remaining area of the plot.
1. Taking A as origin, find the coordinates of P
Answer: a) (4,6)
2. What will be the coordinates of R, if C is the origin?
Answer: c) (10,3)
3. What will be the coordinates of Q, if C is the origin?
b) b) (-6,13)
Answer: d) (13,6)
4. Calculate the area of the triangles if A is the origin
Answer: a) 4.5
5. Calculate the area of the triangles if C is the origin
Answer: d) 4.5
Hope the information shed above regarding Case Study and Passage Based Questions for Class 10 Maths Chapter 7 Coordinate Geometry with Answers Pdf free download has been useful to an extent. If you have any other queries about CBSE Class 10 Maths Coordinate Geometry Case Study and Passage Based Questions with Answers, feel free to comment below so that we can revert back to us at the earliest possible By Team Study Rate
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Case Study Class 10 Maths Questions and Answers (Download PDF)
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Case Study Class 10 Maths
If you are looking for the CBSE Case Study class 10 Maths in PDF, then you are in the right place. CBSE 10th Class Case Study for the Maths Subject is available here on this website. These Case studies can help the students to solve the different types of questions that are based on the case study or passage.
CBSE Board will be asking case study questions based on Maths subjects in the upcoming board exams. Thus, it becomes an essential resource to study.
The Case Study Class 10 Maths Questions cover a wide range of chapters from the subject. Students willing to score good marks in their board exams can use it to practice questions during the exam preparation. The questions are highly interactive and it allows students to use their thoughts and skills to solve the given Case study questions.
Download Class 10 Maths Case Study Questions and Answers PDF (Passage Based)
Download links of class 10 Maths Case Study questions and answers pdf is given on this website. Students can download them for free of cost because it is going to help them to practice a variety of questions from the exam perspective.
Case Study questions class 10 Maths include all chapters wise questions. A few passages are given in the case study PDF of Maths. Students can download them to read and solve the relevant questions that are given in the passage.
Students are advised to access Case Study questions class 10 Maths CBSE chapter wise PDF and learn how to easily solve questions. For gaining the basic knowledge students can refer to the NCERT Class 10th Textbooks. After gaining the basic information students can easily solve the Case Study class 10 Maths questions.
Case Study Questions Class 10 Maths Chapter 1 Real Numbers
Case Study Questions Class 10 Maths Chapter 2 Polynomials
Case Study Questions Class 10 Maths Chapter 3 Pair of Equations in Two Variables
Case Study Questions Class 10 Maths Chapter 4 Quadratic Equations
Case Study Questions Class 10 Maths Chapter 5 Arithmetic Progressions
Case Study Questions Class 10 Maths Chapter 6 Triangles
Case Study Questions Class 10 Maths Chapter 7 Coordinate Geometry
Case Study Questions Class 10 Maths Chapter 8. Introduction to Trigonometry
Case Study Questions Class 10 Maths Chapter 9 Some Applications of Trigonometry
Case Study Questions Class 10 Maths Chapter 10 Circles
Case Study Questions Class 10 Maths Chapter 12 Areas Related to Circles
Case Study Questions Class 10 Maths Chapter 13 Surface Areas & Volumes
Case Study Questions Class 10 Maths Chapter 14 Statistics
Case Study Questions Class 10 Maths Chapter 15 Probability
How to Solve Case Study Based Questions Class 10 Maths?
In order to solve the Case Study Based Questions Class 10 Maths students are needed to observe or analyse the given information or data. Students willing to solve Case Study Based Questions are required to read the passage carefully and then solve them.
While solving the class 10 Maths Case Study questions, the ideal way is to highlight the key information or given data. Because, later it will ease them to write the final answers.
Case Study class 10 Maths consists of 4 to 5 questions that should be answered in MCQ manner. While answering the MCQs of Case Study, students are required to read the paragraph as they can get some clue in between related to the topics discussed.
Also, before solving the Case study type questions it is ideal to use the CBSE Syllabus to brush up the previous learnings.
Features Of Class 10 Maths Case Study Questions And Answers Pdf
Students referring to the Class 10 Maths Case Study Questions And Answers Pdf from Selfstudys will find these features:-
- Accurate answers of all the Case-based questions given in the PDF.
- Case Study class 10 Maths solutions are prepared by subject experts referring to the CBSE Syllabus of class 10.
- Free to download in Portable Document Format (PDF) so that students can study without having access to the internet.
Benefits of Using CBSE Class 10 Maths Case Study Questions and Answers
Since, CBSE Class 10 Maths Case Study Questions and Answers are prepared by our maths experts referring to the CBSE Class 10 Syllabus, it provided benefits in various way:-
- Case study class 10 maths helps in exam preparation since, CBSE Class 10 Question Papers contain case-based questions.
- It allows students to utilise their learning to solve real life problems.
- Solving case study questions class 10 maths helps students in developing their observation skills.
- Those students who solve Case Study Class 10 Maths on a regular basis become extremely good at answering normal formula based maths questions.
- By using class 10 Maths Case Study questions and answers pdf, students focus more on Selfstudys instead of wasting their valuable time.
- With the help of given solutions students learn to solve all Case Study questions class 10 Maths CBSE chapter wise pdf regardless of its difficulty level.
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NCERT Solutions for Class 10 Maths Chapter 7 Free PDF Download
Ncert solutions for class 10 maths chapter 7 – coordinate geometry.
NCERT Solutions for class 10 maths chapter 7 – Coordinate geometry will help you to make your foundation strong on the concepts of Coordinate geometry class 10. The study of Coordinate geometry class 10 and solving the problems will help you to solve complex problems easily. Coordinate geometry class 10 covers all the exercises provided in the NCERT textbook.
CBSE Class 10 Maths Chapter 7 NCERT Solutions are prepared by our expert at Toppr to help you to prepare for your exams in a better way and enhance your score. Coordinate geometry class 10 provide step by step solutions for the questions given in class 10 maths NCERT textbook as per CBSE Board guidelines and are also prepared according to the exam pattern. With the Toppr app, you can download NCERT Solutions for class 10 maths chapter 7 for free. In case you have a doubt while you are studying, Coordinate geometry class 10, for this we have a team of teachers who prove live doubt solving sessions only for you.
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CBSE Class 10 Maths Chapter 7 NCERT Solutions
Coordinate geometry class 10 explains the distance between the two points whose coordinates, area of the triangle formed by three given points, coordinates of the point which divides a line segment joining two points in a given ratio, Distance formula, Section formula, Area of a Triangle. Also, the questions are solved with alternative solutions and diagrammatic representation.
Coordinate Geometry Sub-topics
- 7.1 – Introduction to Coordinate Geometry
- 7.2 – Distance Formula
- 7.3 – Section Formula
- 7.4 – Area of a Triangle
- 7.5 – Summary
You can download NCERT Solutions for Class 10 Maths Chapter 7 PDF for free by clicking on the button below.
NCERT Solutions for Class 10 Maths Chapter 7
Q.1 Find the distance between the following pairs of points:
(2, 3), (4, 1)
(−5, 7), (−1, 3)
( a , b ), (− a , − b )
Distance between two points ( x 1 , y 1 ) and ( x 2 , y 2 ) is
√ − ( x −−−−−−−−−−−−−−−− 1 − x 2 ) 2 + ( y 1 − y 2 ) − 2 (i) Distance between (2, 3) and (4, 1)
= √ − (2 −−−−−−−−−−−−− − 4 ) 2 + (3 − 1 ) − 2 = √ − (−2 −−−−−−−− (2) 2 + ) − 2 = √ − 4+4 −−− = 8√ = 2 2√
(ii) Distance between (−5, 7) and (−1, 3)
= √ − (−5 −−−−−−−−−−−−−−−−− − (−1) ) 2 + (7 − 3 ) − 2 = √ − (−4 −−−−−−−− (4) 2 + ) − 2 = √ − 16 −−−−− + 16 = √ −− 32= 4 2√
(iii )Distance between ( a , b ) and (− a , − b )
= √ − ( a − −−−−−−−−−−−−−−−−−− (− a ) ) 2 + ( b − (− b ) ) − 2 = √ − (2 a −−−−−−−− ) 2 + (2 b ) − 2 = √ − 4 −−−−−−− a 2 + 4 b 2 = 2 √ − a −−−−− 2 + b 2 #465323 Topic: Distance Between Two Points
Q.2 Find the distance between the points (0, 0) and (36, 15) .
Distance Between two given point=
√ − ( x −−−−−−−−−−−−−−−− 2 − x 1 ) 2 + ( y 2 − y 1 ) − 2 Here
x 1 = 0, x 2 = 26 and y 1 = 0, y 2 = 15 ∴ Distance between the points (0, 0) and (36, 15) =
√ − (36 −−−−−−−−−−−−−− − 0 ) 2 + (15 − 0 ) − 2 = √ − +36 −−−−−−− 2 15 2 = √ − 1296 −−−−−−− + 225 − = √ − 1521 −−− = 39
Q.3 Determine if the points (1, 5), (2, 3) and (−2, −11) are collinear.
Three points A , B and C
are collinear if
AB + BC = AC
Here, point A (1, 5), B (2, 3) and
C (−2, −11). ∴ AB = √ − (2 −−−−−−−−−−−−− − 1 ) 2 + (3 − 5 ) − 2 = √ − 1 −−−−−−− 2 +(− 2 2 − ) = √ − 1+4 −−− = 5√ = 2.23 BC = √ − ((−2) −−−−−−−−−−−−−−−−−−−−−− − (2) ) 2 + ((−11) − (3) ) − 2 = √ − (−4 −−−−−−−−−− ) 2 + (−14 ) − 2 = √ − 16 −−−−−− + 196 = √ −−− 212 = 14.56 AC = √ − ((−2) −−−−−−−−−−−−−−−−−−−−−− − (1) ) 2 + ((−11) + (5) ) − 2 = ( √ − 3 ) 2 + (−16 ) 2 = √ − 9 −−−−− + 256 = √ −−− 265 = 16.27 AB + BC = 2.23 + 14.56 = 16.79
AB + BC ≃ AC
Hence the given points are collinear.
Q. 4 Find the ratio in which the line segment joining the points (−3, 10) and (6, −8) is divided by (−1, 6)
Let the required ratio be
Take ( x 1 , y 1 ) = (−3, 10); ( x 2 , y 2 ) = (6, −8) and
( x , y ) = (−1, 6) ∴ ⇒ x −1 = = m 1 xm m 2 1 1 + m 2 x 1
m 2 ×−3 m 1 + m 2 ⇒− m 1 − m 2 =6 m − 3 m 2
⇒− m 1 −6 m 1 =−3 m 2 + m 2
⇒ −7 m 1 = −2 m 2 ⇒ m 1 m 2
= 2 7 ∴ The required ratio is 2:7
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NCERT Solutions for Class 10
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NCERT Solutions for Class 10 Maths Chapter 7- Coordinate Geometry
All of the tasks in the NCERT textbook for Class 10 Maths Chapter 7 Coordinate Geometry are covered here in the NCERT Solutions. All the Textbook questions are solved here in this section with detailed explanation of the basic concepts used in here by our subject specialists.
All the solutions are solved in such a way to make these concepts very much understandable for the students so that they can do any kind of questions related to the topic.
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If students go through the NCERT solutions thoroughly, it makes them understand the fundamentals of the chapter which is very helpful for further exams as well as higher classes.
All these solutions are prepared by our highly qualified and experienced specialists. The solutions are evaluated by our professions as well to ensure there’s no doubt of error and each and every solution is in easy language for students who are studying the chapter for the very first time.
NCERT Solutions for Class 10 Maths Chapter 7- Coordinate Geometry | Free PDF Download
NCERT Solutions are solved in student friendly way to assist students in achieving high scores in the examinations. It gives you much-needed problem-solving experience.
For easy access of the solutions of the questions of this chapter, the chapter is solved in different sections and each section contains the exercise wise solutions.
As the Chapter name suggests, Coordinate geometry , this chapter talks “{ about the geometry by using coordinates in a plane.
Coordinate plane is basically a plane formed with two perpendicular lines, the horizontal line is known as x -axis and the vertical line is known as the y-axis. The value of x-axis is known as abscissa and the value of y axis is known as the ordinate. These sets of values of abscissa and ordinate help us to show the exact value of the position of a point in a coordinate plane.
Exercise 7.1 is about how to find the distance between two points. The formula thus used is called Distance Formula .
Exercise 7.2 explains the Section Formula . Let us suppose a line segment AB in which coordinates of A and B are known and a point P divides AB in a particular ratio, say, m:n, then the coordinates of point can be found using Section Formula .
Exercise 7.3 includes the formula to find the area of a triangle if coordinates of the vertices are known.
Exercise 7.4 covers all the three concepts learned in the previous three exercises which increases the understanding of this chapter to a whole new level.
These NCERT Solutions for Class 10 Maths include a wide range of questions and answers to help you with alternative solutions and representation of solutions with diagrams. The solutions are presented in common English and include important information. If students study these NCERT Solutions thoroughly, they will be able to solve more complex questions with ease.
This chapter has weightage of 6 marks in the CBSE board examinations and much more in other higher exams.
NCERT Solutions for Class 10 Maths Chapter 7- Co-ordinate geometry: Key Benefits
- All of the exercise questions from the NCERT textbook are included in these solutions.
- This NCERT Solution contains a variety of examples that will assist you in relating geometry and numerical concepts to real-life situations.
- Alternative approaches and diagrammatic representations will assist you in fully comprehending the subject.
- You will become familiar with important formulas and standards by solving these NCERT Solutions.
- It follows the NCERT guidelines, which assist students in their preparation to score good in the examination and boost confidence.
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NCERT Solutions Class 10 Maths All Chapters:
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Frequently Asked Questions: NCERT Solutions of Class 10 Maths Chapter 7- Coordinate Geometry
Are NCERT solutions for class 10 Maths chapter 7 helpful from examination point of view?
Answer: Yes, NCERT solutions for class 10 Maths chapter 7 are very much helpful from examination point of view. Total of 6 marks of questions come in the examination from this chapter. This chapter is very much easier to understand and thus these 6 marks can be scored very easily.
Is it okay to just solve the example problems before the exercise of chapter?
Answer: No, it is compulsory to solve all the questions of the exercises of chapter 7 Coordinate Geometry as solving all the questions in the exercise will give you a clear understanding of the chapter and will build your confidence for the examination. Solving all the questions will help you memorize the different formulas of the chapter without much effort.
What are the formulas included in the Class 10 Maths Chapter 7 Coordinate Geometry?
Answer: Class 10 Maths Chapter 7 Coordinate Geometry covers the distance formula, to find distance between two points; section formula, to find the coordinates of the point dividing a line segment in a particular ratio; and the formula to find the area of the triangle when the coordinates of vertices of triangle are known. Students will be able to tackle complex problems related to the chapter with ease if they go through these NCERT solutions thoroughly.
Is it necessary to solve NCERT problems?
Answer: Yes, answering NCERT questions is important because it clarifies the fundamentals of the questions, which will be useful even if the examiner distorts the wording or pattern of the core questions.
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Most Important Questions Class 10 Maths Coordinate Geometry with Solutions
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Mathematics Most Important Questions of Coordinate Geometry Class 10 contains various questions that are useful for students even in their competitive examinations like JEE & NEET. The Coordinate Geometry chapter is very beneficial for innovative people. The photocopy machine gives the exact copy of the papers by using the coordinate geometry and Air Traffic is also managed by using the Coordinate Geometry.
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In the table given below, we have provided the links to downloadable Class 10 Coordinate Geometry Extra Questions PDFs. Now you can download them without requiring you to login.
| TYPES OF QUESTIONS | IMPORTANT QUESTION PDFs |
|---|---|
| MCQ (1 Mark) | |
| VSA (2 Marks) | |
| SA I (3 Marks) | |
| SA II (4 Marks) |
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NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry
Class 10 Maths NCERT Solutions Chapter 7: NCERT Solutions provided here for Coordinate Geometry covers all the exercises present in NCERT Textbooks. The NCERT Class 10 Maths Multiple Choice Questions prepared by our subject expertise acts as study material for the students preparing for board exams. You can access the notes, study material, formulas and important questions for CBSE Class 10 Maths Chapter 7 covered as per the latest NCERT Syllabus. Students can utilize the NCERT Solutions for Class 10 Maths Pdf of Chapter 7 by downloading them or by bookmarking the site and referring them as and when needed. By practicing on a regular basis you will get perfection on the topics involved in Coordinate Geometry. Get to know the step by step answers provided for various kinds of questions asked in CBSE Maths Class 10 Coordinate Geometry Exercises 7.1, 7.2, 7.3, 7.4 by referring further. CBSE Class 10 Maths Chapter 7 Solutions are prepared as per the UP, MP and CBSE Board Standards.
Topics and SubTopics List in Maths Class 10 Coordinate Geometry
| 7 | Coordinate Geometry |
| 7.1 | Introduction |
| 7.2 | Distance Formula |
| 7.3 | Section Formula |
| 7.4 | Area of the Triangle |
| 7.5 | Summary |
NCERT Solutions for Class 10 Maths Chapter 7 – Coordinate Geometry
If you are a student of Class 10 you might have across the Chapter 7 Coordinate Geometry in NCERT Textbooks while preparing for your examination. After going through the questions in Exercises you might be eagerly looking for solutions. You can halt your search as we have come up with Maths NCERT Solutions Class 10 for Coordinate Geometry in an elaborate manner. Keeping in mind the preferences of students we tried providing the CBSE Solutions Class 10 Maths Chapter 7 Coordinate Geometry in both Hindi and English Mediums respectively.
| Class | 10 |
| Subject | Maths |
| Book | Mathematics |
| Chapter No. | 7 |
| Chapter Name | Coordinate Geometry |
Class 10 Maths NCERT Solutions Chapter 7 Coordinate Geometry Ex 7.1
Class 10 Maths NCERT Solutions Chapter 7 Coordinate Geometry Ex 7.2
Class 10 Maths NCERT Solutions Chapter 7 Coordinate Geometry Ex 7.3
Class 10 Maths NCERT Solutions Chapter 7 Coordinate Geometry Ex 7.4
NCERT Class 10 Maths Solutions Chapter 7 – Solved Exercises
Brush up the concepts well before attending the Exam with the NCERT Maths Solutions Class 10 Coordinate Geometry and understand the methods used for approach. By doing so you can attempt the Class 10 Maths Board Examination with confidence and score well. Get to know about the Maths Class 10 Maths Coordinate Geometry Exercises 7.1, 7.2 7.3, 7.4 explained in detail.
Total weightage from Coordinate Geometry in Class 10 Maths Exam is 6 Marks. Out of which there will be a 1 Mark, 2 Mark, and 3 Mark Question. Revise and solve a number of Exemplar Problems, Important Questions with the CBSE Maths Class 10 Solutions provided by us. You can view them either online or download the PDF and use it for further reference.
We gave you the Maths Class 10 NCERT Solutions Chapter 7 Coordinate Geometry as per the Latest CBSE Syllabus and Study materials revised from time to time. Make use of the Formulas, Notes, Study Material provided for Class 10 Maths Examination and score good marks.
We wish the information shared above related to Class 10 Maths NCERT Solutions Chapter 7 gave you little idea on what will be there in it. In case of any doubts feel free to reach us via comment section and we will be of your help.
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Class 10 Maths Extra Questions Chapter 7 Coordinate Geometry
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QB365 Provides the updated CASE Study Questions for Class 10 Maths, and also provide the detail solution for each and every case study questions . Case study questions are latest updated question pattern from NCERT, QB365 will helps to get more marks in Exams
CBSE Class 10 Maths Case Study Questions for Chapter 7
Hope the information shed above regarding Case Study and Passage Based Questions for Class 10 Maths Chapter 7 Coordinate Geometry with Answers Pdf free download has been useful to an extent. If you have any other queries about CBSE Class 10 Maths Coordinate Geometry Case Study and Passage Based Questions with Answers, feel free to comment below ...
Case Study on Coordinate Geometry Class 10 Maths PDF
Mere Bacchon, you must practice the CBSE Case Study Questions Class 10 Maths Coordinate Geometry in order to fully complete your preparation.They are very very important from exam point of view. These tricky Case Study Based Questions can act as a villain in your heroic exams!. I have made sure the questions (along with the solutions) prepare you fully for the upcoming exams.
Case Study Questions: Question 1: The top of a table is shown in the figure given below: (i) The coordinates of the points H and G are respectively(a) (1, 5), (5, 1) (b) (0, 5), (5, 0) (c) (1, 5), (5, 0) (d) (5, 1), (1, 5) (ii) The distance between the points A and … Continue reading Case Study Questions for Class 10 Maths Chapter 7 Coordinate Geometry
These Case Study and Passage Based questions are published by the experts of CBSE Experts for the students of CBSE Class 10 so that they can score 100% on Boards. CBSE Class 10 Mathematics Exam 2024 will have a set of questions based on case studies in the form of MCQs. The CBSE Class 10 Mathematics Question Bank on Case Studies, provided in ...
The Case Based Questions: Coordinate Geometry is an invaluable resource that delves deep into the core of the Class 10 exam. These study notes are curated by experts and cover all the essential topics and concepts, making your preparation more efficient and effective.
Class 10th Maths - Coordinate Geometry Case Study Questions and Answers 2022 - 2023 - Complete list of 10th Standard CBSE question papers, syllabus, exam tips, study material, previous year exam question papers, centum tips, formula, answer keys, solutions etc.. ... QB365 provides a detailed and simple solution for every Possible Case Study ...
COORDINATE GEOMETRY CHAPTER 7 (A) Main Concepts and Results Distance Formula, Section Formula, Area of a Triangle. • The distance between two points P (x 1, y 1) and Q (x2, y 2) is ( )( )2 2 xx y y21 2 1-- + • The distance of a point P (x,y) from the origin is xy22+ • The coordinates of the point P which divides the line segment joining the points
Now that you have gone through the questions, download volume 3 and 4 of CBSE Class 10 Mathematics Chapter 7 Coordinate Geometry competency-based practice questions from the links below.
Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now. Case Study on Coordinate Geometry Class 10 Maths: Here, you will get Case Study Questions on Class 10 Coordinate Geometry PDF at free of cost. Along with you can also download Coordinate Geometry case study questions for ...
Case Study Class 10 Maths Questions
Coordinate Geometry Class 10 Maths Important Questions Very Short Answer (1 Mark) 1. The ordinate of a point A on y-axis is 5 and B has co-ordinates (-3, 1). Find the length of AB. Solution. We have, A (0, 5) and B (-3, 1). Distance between A and B, 2.
Coordinate Geometry
RD Sharma Solutions Class 10 Maths Chapter 14 - Free PDF Download. The RD Sharma Solutions for Class 10 Maths Chapter 14 - Co-ordinate Geometry explain each concept in a comprehensive manner to help students obtain a thorough knowledge of concepts. Coordinates are ordered pairs of numbers, and the study of Geometry using coordinates is called Co-ordinate Geometry.
Here, we have provided case-based/passage-based questions for Class 10 Maths Chapter 7 Coordinate Geometry. Case Study/Passage-Based Questions. Question 1: A satellite image of a colony is shown below. In this view, a particular house is pointed out by a flag, which is situated at the point intersection of the x and y-axes.
Case Study Class 10 Maths Questions and Answers ...
Coordinate geometry class 10 provide step by step solutions for the questions given in class 10 maths NCERT textbook as per CBSE Board guidelines and are also prepared according to the exam pattern. With the Toppr app, you can download NCERT Solutions for class 10 maths chapter 7 for free.
What are the formulas included in the Class 10 Maths Chapter 7 Coordinate Geometry? Answer: Class 10 Maths Chapter 7 Coordinate Geometry covers the distance formula, to find distance between two points; section formula, to find the coordinates of the point dividing a line segment in a particular ratio; and the formula to find the area of the ...
Mathematics Most Important Questions of Coordinate Geometry Class 10 contains various questions that are useful for students even in their competitive examinations like JEE & NEET. The Coordinate Geometry chapter is very beneficial for innovative people. The photocopy machine gives the exact copy of the papers by using the coordinate geometry and Air Traffic is also managed by using the ...
Class 10 Maths NCERT Solutions Chapter 7: NCERT Solutions provided here for Coordinate Geometry covers all the exercises present in NCERT Textbooks. The NCERT Class 10 Maths Multiple Choice Questions prepared by our subject expertise acts as study material for the students preparing for board exams. You can access the notes, study material ...
Download Class 10 Maths Extra Questions Chapter 7 Coordinate Geometry free from Aglasem Docs. aglasem.com. Login. ... Class 10 Maths Extra Questions Chapter 7 Coordinate Geometry More Detail. ... It provides all important study material and resources for free to students.
Mathematics document from Delhi Public School - Durg, 4 pages, Coordinate Geometry PYQs YouTube Channel - Ashish4students Ashish4students - YouTube Student's Success is My Real Success Coordinate Geometry PYQs 1. Find the point on − which is equidistant from the points (2, −2) (−4, 2). [CBSE SQP 2020-21] 2. (−2, 5)
The document is a case study assignment from Bhatnagar International School for Class X Mathematics. It provides two case studies related to coordinate geometry. The first case study involves archaeologists plotting the locations of artifacts found at an excavation site on a coordinate plane. It then asks students multiple choice questions about the distances and coordinates of a medicine ...