**NCERT Solutions for class 10 Chapter 7 Coordinate Geometry** covers all the exercise provided in NCERT textbook. These** NCERT solutions,** prepared by experts at BYJU’S, are a comprehensive study material for the students preparing for CBSE class 10 board examination. These solutions are available for easy access and download of the students. Here you can get detail stepwise answers to different types of questions provided in the NCERT textbook. Practicing the solutions of NCERT will help you attain perfection on the topics involved in coordinate geometry chapter.

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### Class 10 Maths Chapter 7 Exercise7.1 Page No: 161

**1. Find the distance between the following pairs of points: **

**(i) (2, 3), (4, 1) **

**(ii) (-5, 7), (-1, 3) **

**(iii) (a, b), (- a, – b) **

**2. Find the distance between the points (0, 0) and (36, 15). Can you now find the distance between the two towns A and B discussed in Section 7.2. **

**3. Determine if the points (1, 5), (2, 3) and (-2, -11) are collinear.**

**4. Check whether (5, – 2), (6, 4) and (7, – 2) are the vertices of an isosceles triangle. **

**Solution:**

**5. In a classroom, 4 friends are seated at the points A, B, C and D as shown in Fig. 7.8. Champa and Chameli walk into the class and after observing for a few minutes Champa asks Chameli, “Don’t you think ABCD is a square?” Chameli disagrees. Using distance formula, find which of them is correct.**

**6. Name the type of quadrilateral formed, if any, by the following points, and give reasons for your answer: **

**(i) (- 1, – 2), (1, 0), (- 1, 2), (- 3, 0)**

** (ii) (- 3, 5), (3, 1), (0, 3), (- 1, – 4) **

**(iii) (4, 5), (7, 6), (4, 3), (1, 2)**

Opposite sides of this quadrilateral are of the same length. However, the diagonals are of different lengths. Therefore, the given points are the vertices of a parallelogram.

**7. Find the point on the x-axis which is equidistant from (2, – 5) and (- 2, 9). **

**8. Find the values of y for which the distance between the points P (2, – 3) and Q (10, y) is 10 units.**

**Solution:**

Given: Distance between (2, – 3) and (10, y) is 10.

Using distance formula,

**9. If Q (0, 1) is equidistant from P (5, – 3) and R (x, 6), find the values of x. Also find the distance QR and PR. **

**10. Find a relation between x and y such that the point (x, y) is equidistant from the point (3, 6) and (- 3, 4). **

### Class 10 Maths Chapter 7 Exercise 7.2 Page No: 167

**1. Find the coordinates of the point which divides the join of (- 1, 7) and (4, – 3) in the ratio 2:3. **

**Solution:**

**2. Find the coordinates of the points of trisection of the line segment joining (4, -1) and (-2, -3). **

**Solution:**

Let P (*x*1, *y*1) and Q (*x*2, *y*2) are the points of trisection of the line segment joining the given points i.e., AP = PQ = QB

Therefore, point P divides AB internally in the ratio 1:2.

**3. To conduct Sports Day activities, in your rectangular shaped school ground ABCD, lines have been drawn with chalk powder at a distance of 1 m each. 100 flower pots have been placed at a distance of 1 m from each other along AD, as shown in the following figure. Niharika runs 1/4 th the distance AD on the 2nd line and posts a green flag. Preet runs 1/5th the distance AD on the eighth line and posts a red flag. What is the distance between both the flags? If Rashmi has to post a blue flag exactly halfway between the line segment joining the two flags, where should she post her flag?**

**Solution:**

From the given instruction, we observed that Niharika posted the green flag at 1/4th of the distance AD i.e., (1/4 ×100)m = 25m from the starting point of 2nd line. Therefore, the coordinates of this point are (2, 25).

Similarly, Preet posted red flag at 1/5 of the distance AD i.e., (1/5 ×100) m = 20m from the starting point of 8th line. Therefore, the coordinates of this point are (8, 20).

Distance between these flags can be calculated by using distance formula,

**4. Find the ratio in which the line segment joining the points (-3, 10) and (6, – 8) is divided by (-1, 6).**

**Solution:**

Consider the ratio in which the line segment joining ( -3, 10) and (6, -8) is divided by point ( -1, 6) be k :1.

Therefore, -1 = ( 6*k*-3)/(*k*+1)

–*k *– 1 = 6*k *-3

7*k *= 2

*k *= 2/7

Therefore, the required ratio is 2:7.

**5. Find the ratio in which the line segment joining A (1, – 5) and B (- 4, 5) is divided by the x-axis. Also find the coordinates of the point of division. **

**6. If (1, 2), (4, y), (x, 6) and (3, 5) are the vertices of a parallelogram taken in order, find x and y.**

**7. Find the coordinates of a point A, where AB is the diameter of circle whose centre is (2, – 3) and B is (1,4).**

**8. If A and B are (-2, -2) and (2, -4), respectively, find the coordinates of P such that AP = 3/7 AB and P lies on the line segment AB. **

**9. Find the coordinates of the points which divide the line segment joining A (- 2, 2) and B (2, 8) into four equal parts. **

**10. Find the area of a rhombus if its vertices are (3, 0), (4, 5), (-1, 4) and (-2,-1) taken in order.**

**[Hint: Area of a rhombus = 1/2(product of its diagonals)]**

### Class 10 Maths Chapter 7 Exercise 7.3 Page No: 170

**1. Find the area of the triangle whose vertices are: **

**(i) (2, 3), (-1, 0), (2, -4) **

**(ii) (-5, -1), (3, -5), (5, 2)**

**2. In each of the following find the value of ‘k’, for which the points are collinear.**

** (i) (7, -2), (5, 1), (3, -k) **

**(ii) (8, 1), (k, -4), (2, -5) **

**Solution:**

- For collinear points, area of triangle formed by them is always zero.

Let points (7, -2) (5, 1), and (3, k) are vertices of a triangle.

Area of triangle = 1/2 [7 { 1- k} + 5(k-(-2)) + 3{(-2) – 1}] = 0

7 – 7k + 5k +10 -9 = 0

-2k + 8 = 0

k = 4

(ii) For collinear points, area of triangle formed by them is zero.

Therefore, for points (8, 1), (k, – 4), and (2, – 5), area = 0

1/2 [8 { -4- (-5)} + k{(-5)-(1)} + 2{1 -(-4)}] = 0

8 – 6k + 10 = 0

6k = 18

k = 3

**3. Find the area of the triangle formed by joining the mid-points of the sides of the triangle whose vertices are (0, -1), (2, 1) and (0, 3). Find the ratio of this area to the area of the given triangle. **

**4. Find the area of the quadrilateral whose vertices, taken in order, are**

** (-4, -2), (-3, -5), (3, -2) and (2, 3).**

**5. You have studied in Class IX that a median of a triangle divides it into two triangles of equal areas. Verify this result for ΔABC whose vertices are A (4, – 6), B (3, – 2) and C (5, 2).**

### Class 10 Maths Chapter 7 Exercise 7.4 Page No: 171

**1. Determine the ratio in which the line 2x + y – 4 = 0 divides the line segment joining the points A(2, –2) and B(3, 7). **

**2. Find the relation between x and y if the points (x, y), (1, 2) and (7, 0) are collinear. **

**Solution:**

If given points are collinear then area of triangle formed by them must be zero.

Let (x, y), (1, 2) and (7, 0) are vertices of a triangle,

AREA OF A TRIANGLE = ½ (x1(y2 – y3 ) + x2(y3 – y1) + x3(y1 – y2) ) = 0

= 1/2[x(2 – 0) + 1 (0 – y) + 7( y – 2)] = 0

- 2x – y + 7y – 14 = 0
- 2x + 6y – 14 = 0
- x + 3y – 7 = 0. Which is required result.

**3. Find the centre of a circle passing through points (6, -6), (3, -7) and (3, 3).**

**4. The two opposite vertices of a square are (-1, 2) and (3, 2). Find the coordinates of the other two vertices.**

**5. The class X students of a secondary school in Krishinagar have 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 lawn in the plot as shown in the fig. 7.14. The students are to sow the seeds of flowering plants on the remaining area of the plot. **

**(i) Taking A as origin, find the coordinates of the vertices of the triangle. **

**(ii) What will be the coordinates of the vertices of triangle PQR if C is the origin?**

**Also calculate the areas of the triangles in these cases. What do you observe? **

**Solution: **

(i) Taking A as origin, coordinates of the vertices P, Q and R are,

From figure: P = (4, 6), Q = (3, 2), R (6, 5)

Here AD is the x-axis and AB is the y-axis.

(ii) Taking C as origin,

Coordinates of vertices P, Q and R are ( 12, 2) , (13, 6) and (10, 3) respectively.

Here CB is the x-axis and CD is the y-axis.

Find the area of triangles:

Area of triangle PQR in case of origin A:

Using formula: Area of a triangle =

= ½ [4(2 – 5) + 3 (5 – 6) + 6 (6 – 2)]

= ½ ( – 12 – 3 + 24 )

= 9/2 sq unit

(ii) Area of triangle PQR in case of origin C:

Area of a triangle =

= ½ [ 12(6 – 3) + 13 ( 3 – 2) + 10( 2 – 6)]

= ½ ( 36 + 13 – 40)

= 9/2 sq unit

This implies, Area of triangle PQR at origin A = Area of triangle PQR at origin C

Area is same in both case because triangle remains the same no matter which point is considered as origin.

**6. The vertices of a ∆ ABC are A (4, 6), B (1, 5) and C (7, 2). A line is drawn to intersect sides AB and AC at D and E respectively, such that AD/AB = AE/AC = ¼ . Calculate the area of the ∆ ADE and compare it with area of ∆ ABC. (Recall Theorem 6.2 and Theorem 6.6)**

**7. Let A (4, 2), B (6, 5) and C (1, 4) be the vertices of ∆ ABC. **

**(i) The median from A meets BC at D. Find the coordinates of point D. **

**(ii) Find the coordinates of the point P on AD such that AP : PD = 2 : 1. **

**(iii) Find the coordinates of point Q and R on medians BE and CF respectively such that BQ : QE = 2:1 and CR : RF = 2 : 1.**

**(iv) What do you observe? **

**[Note : The point which is common to all the three medians is called the centroid**

**and this point divides each median in the ratio 2 : 1.]**

**(v) If A (x _{1}, y_{1}), B (x_{2}, y_{2}) and C (x_{3}, y_{3}) are the vertices of triangle ABC, find the coordinates of the centroid of the triangle.**

**8. ABCD is a rectangle formed by the points A (-1, – 1), B (-1, 4), C (5, 4) and D (5, -1). P, Q, R and S are the midpoints of AB, BC, CD and DA respectively. Is the quadrilateral PQRS a square? a rectangle? or a rhombus? Justify your answer.**

**Solution:**

## NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry

This chapter comes under unit 6 and has a weightage of 6 marks in the board examination. There will be one mark MCQ question, 2mark reasoning questions and 3 marks short answer questions. This chapter has fundamental concepts that lay the foundation for your future studies.

### Sub-topics of class 10 chapter 7 Coordinate Geometry

- Introduction to Coordinate Geometry
- Distance Formula
- Section Formula
- Area of the Triangle

### List of Exercise from class 10 Maths Chapter 7 Coordinate Geometry

Exercise 7.1– 10 Questions which includes 8 practical based questions, 2 reasoning questions

Exercise 7.2– 10 Questions which includes 8 long answer questions, 2 short answer questions

Exercise 7.3– 5 Questions which includes 3 long answer questions, 2 practical based questions

Exercise 7.4– 8 Questions which includes 6 long answer questions, 1 practical based question, 1 reasoning question

NCERT Solutions are carefully drafted to assist the student in scoring good marks in the examination. It provides you with much-needed problem-solving practice.

This chapter deals with finding the area between two points whose coordinate values are provided. For instance area of triangle. This chapter has some basic concepts like area of triangle, rhombus, distance between sides and intersections. This chapter teaches you the relationship between numerical and geometry and their application in our daily lives.

These NCERT Solution for Class 10 Maths have different types of questions and their answers which will help you with alternate solutions, diagrammatic representation. Solution provided here is written in a simple language with apt information. By studying this NCERT solution thoroughly students will be able to solve complex problems easily.

### Key Features of NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry

- The solution has to all the exercise questions provided in the NCERT textbook
- Diagrammatic representations and alternate methods will help you understand the concepts thoroughly.
- Solving this NCERT Solution will make you acquainted with important formulas and standards.
- This NCERT Solution has different examples that will make you relate to real-life examples to relate geometry and numerical.

BYJU’S is India’s top online education provider with countries best teachers on its board. You can get NCERT study materials on BYJU’S site and application. To get access to all the study materials visit BYJU’S website or download BYJU’S learning App.

Also Access |

NCERT Exemplar for class 10 Maths Chapter 7 Coordinate Geometry |

CBSE Notes for Class 10 Maths Chapter 7 Coordinate Geometry |

## Frequently Asked Questions on Chapter 7 Coordinate Geometry

### Find the coordinates of the point which divides the join of (- 1, 7) and (4, – 3) in the ratio 2:3

Let P(x,y) be the required point.

By using the section formula,

x = (2×4 + 3×(-1))/(2+3)

= (8-3)/5

= 1

y = (2×(-3) + 3×7)/(2+3)

= (-6+21)/5

= 3

Hence, the point is (1,3).

### Find the ratio in which the line segment joining the points (-3, 10) and (6, – 8) is divided by (-1, 6)

Consider the ratio in which the line segment joining ( -3, 10) and (6, -8) is divided by point ( -1, 6) be k :1.

-1 = ( 6k-3)/(k+1)

–k – 1 = 6k -3

7k = 2

k = 2/7

Hence, the required ratio is 2:7.

### In the given points (7, -2), (5, 1), (3, -k) find the value of ‘k’, for which the points are collinear.

For collinear points, the area of the triangle formed by them is always zero.

So, let the points (7, -2) (5, 1), and (3, k) be the vertices of a triangle.

Area of triangle = 1/2 [7 { 1- k} + 5(k-(-2)) + 3{(-2) – 1}] = 0

7 – 7k + 5k +10 -9 = 0

-2k + 8 = 0

k = 4

Hence, the value of k is 4.