# NCERT Solutions Class 10 for Maths Chapter 7 Coordinate Geometry

NCERT Solutions Class 10 for **Maths Chapter 7 Coordinate Geometry** : In this post, we will share with you all the detailed **NCERT Solutions of Class 10 Maths Chapter 7 Coordinate Geometry**. This will contain both in-text and back-exercise questions for Science and Social Science, and all exercise questions for Mathematics. For all school and board level examinations, doing all the NCERT Questions is a must.

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# Class 10 Maths Chapter 7 – Coordinate Geometry

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**Exercise 7.1 (Page 161)**

**Question 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)

**Answer 1:**

**Question 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.

**Answer 2:**

**Question 3**:

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

**Answer 3:**

**Question 4:**

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

**Answer 4:**

**Question 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.

**Answer 5:**

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**Question 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)

**Answer 6:**

**Question 7:**

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

**Answer 7:**

**Question 8:**

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

**Answer 8:**

**Question 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.

**Answer 9:**

**Question 10:**

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

**Answer 10:**

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## Exercise 7.2 (Page 167)

**Question 1:**

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

**Answer 1:**

Let P(x, y) be the required point. Using the section formula, we get

x = (2×4+3×(-1))/(2+3) = (8-3)/5 = 1

y = (2×-3+3×7)/(2+3) = (-6+21)/5 = 3

Therefore, the point is (1, 3).

**Question 2:**

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

**Answer 2:**

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.

x_{1 }= (1×(-2)+2×4)/3 = (-2+8)/3 = 6/3 = 2

y_{1} = (1×(-3)+2×(-1))/(1+2) = (-3-2)/3 = -5/3

Therefore: P (*x*1, *y*1) = P(2, -5/3)

Point Q divides AB internally in the ratio 2:1.

x_{2} = (2×(-2)+1×4)/(2+1) = (-4+4)/3 = 0

y_{2} = (2×(-3)+1×(-1))/(2+1) = (-6-1)/3 = -7/3

The coordinates of the point Q is (0, -7/3)

**Question 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?

**Answer 3:**

**Question 4**:

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

**Answer 4:**

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.

**Question 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.

**Answer 5:**

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**Question 6:**

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

**Answer 6:**

Let A,B,C and D be the points of a parallelogram : A(1,2), B(4,y), C(x,6) and D(3,5).

Since the diagonals of a parallelogram bisect each other, the midpoint is same.

To find the value of x and y, solve for midpoint first.

Midpoint of AC = ( (1+x)/2 , (2+6)/2 ) = ((1+x)/2 , 4)

Midpoint of BD = ((4+3)/2 , (5+y)/2 ) = (7/2 , (5+y)/2)

Midpoint of AC and BD are same, this implies

(1+x)/2 = 7/2 and 4 = (5+y)/2^{x }^{+ 1 = 7 and 5 + }^{y }^{= 8}*x *= 6 and *y *= 3.

**Question 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).

**Answer 7:**

Let the coordinates of point A be (x, y).

Mid-point of AB is (2, – 3), which is the centre of the circle.

Coordinate of B = (1, 4)

(2, -3) =((x+1)/2 , (y+4)/2)

(x+1)/2 = 2 and (y+4)/2 = -3

x + 1 = 4 and y + 4 = -6

x = 3 and y = -10

The coordinates of A(3,-10). Answer!

**Question 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.

**Answer 8:**

**Question 9:**

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

**Answer 9:**

**Question 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)

**Answer 10:**

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## Exercise 7.3 (Page 170)

**Question 1:**

Find the area of the triangle whose vertices are:

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

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

**Answer 1:**

(i) Here,

Substitute all the values in the above formula, we get

Area of triangle = 1/2 [2 {0- (-4)} + (-1) {(-4) – (3)} + 2 (3 – 0)]

= 1/2 {8 + 7 + 6} = 21/2

So area of triangle is 21/2 square units.

(ii) Here,

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

= 1/2{35 + 9 + 20} = 32

Therefore, area of the triangle is 32 square units.

**Question 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)

**Answer 2:**

(i) 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

**Question 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.

**Answer 3:**

Let the vertices of the triangle be A (0, -1), B (2, 1), C (0, 3).

Let D, E, F be the mid-points of the sides of this triangle.

Coordinates of D, E, and F are given by

D = ( , ) = (1, 0)

E = ( , ) = (0, 1)

F = ( , ) = (1, 2)

Area of a triangle =

Area of ΔDEF = 1/2 {1(2-1) + 1(1-0) + 0(0-2)} = 1/2 (1+1) = 1

Area of ΔDEF is 1 square units

Area of ΔABC = 1/2 [0(1-3) + 2{3-(-1)} + 0(-1-1)] = 1/2 {8} = 4

Area of ΔABC is 4 square units

Therefore, the required ratio is 1:4.

**Question 4:**

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

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

**Answer 4:**

Let the vertices of the quadrilateral be A ( – 4, – 2), B ( – 3, – 5), C (3, – 2), and D (2, 3). Join AC to form two triangles ΔABC and ΔACD.

Area of a triangle = 1/2 {*x*_{1} (*y*_{2} – *y*_{3}) + *x*_{2} (*y*_{3} – *y*_{1}) + *x*_{3} (*y*_{1} – *y*_{2})}

Area of ΔABC = 1/2 [(-4) {(-5) – (-2)} + (-3) {(-2) – (-2)} + 3 {(-2) – (-5)}]

= 1/2 (12+0+9)

= 21/2 square units

Area of ΔACD = 1/2 [(-4) {(-2) – (3)} + 3{(3) – (-2)} + 2 {(-2) – (-2)}]

= 1/2 (20+15+0)

= 35/2 square units

Area of ☐ABCD = Area of ΔABC + Area of ΔACD

= (21/2 + 35/2) square units = 28 square units

**Question 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).

**Answer 5:**

Let the vertices of the triangle be A (4, -6), B (3, -2), and C (5, 2)

Let D be the mid-point of side BC of ΔABC. Therefore, AD is the median in ΔABC.

Coordinates of point D = Midpoint of BC = ( = (4, 0)

Formula, to find Area of a triangle =

Now, Area of ΔABD = 1/2 [(4) {(-2) – (0)} + 3{(0) – (-6)} + (4) {(-6) – (-2)}]

= 1/2 (-8 + 18 – 16)

= -3 square units

However, area cannot be negative. Therefore, area of ΔABD is 3 square units.

Area of ΔACD = 1/2 [(4) {0 – (2)} + 4{(2) – (-6)} + (5) {(-6) – (0)}]

= 1/2 (-8 + 32 – 30) = -3 square units

However, area cannot be negative. Therefore, area of ΔACD is 3 square units.

The area of both sides is same. Thus, median AD has divided ΔABC in two triangles of equal areas.

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## Exercise 7.4 (Page 171)

**Question 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).

**Answer 1:**

Consider line 2x + y – 4 = 0 divides line AB joined by the two points A(2, -2) and B(3, 7) in k : 1 ratio.

Coordinates of point of division can be given as follows:

x = and y =

Substituting the values of x and y given equation, i.e. 2x + y – 4 = 0, we have

2( + (– 4 = 0

( = 4

4 + 6k – 2 + 7k = 4(k+1)

-2 + 9k = 0

Or k = 2/9

Hence, the ratio is 2:9.

**Question 2:**

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

**Answer 2:**

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 = = 0

[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.

**Question 3:**

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

**Answer 3:**

Let A = (6, -6), B = (3, -7), C = (3, 3) are the points on a circle.

If O is the centre, then OA = OB = OC (radii are equal)

If O = (x, y) then

OA =

OB =|

OC =

Choose: OA = OB, we have

After simplifying above, we get -6x = 2y – 14 ….(1)

Similarly: OB = OC

(x – 3)2 + (y + 7)2 = (x – 3)2 + (y – 3)2

(y + 7)2 = (y – 3)2

y2 + 14y + 49 = y2 – 6y + 9

20y =-40

or y = -2

Substituting the value of y in equation (1), we get;

-6x = 2y – 14

-6x = -4 – 14 = -18

x = 3

Hence, centre of the circle located at point (3,-2).

**Question 4:**

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

**Answer 4:**

Let ABCD is a square, where A(-1,2) and B(3,2). And Point O is the point of intersection of AC and BD

To Find: Coordinate of points B and D.

Step 1: Find distance between A and C and coordinates of point O.

We know that, diagonals of a square are equal and bisect each other.

AC = = 4

Coordinates of O can be calculated as follows:

x = (3-1)/2 = 1 and y = (2+2)/2 = 2

So O(1,2)

Step 2: Find the side of the square using Pythagoras theorem

Let a be the side of square and AC = 4

From right triangle, ACD,

a = 2√2

Hence, each side of square = 2√2

Step 3: Find coordinates of point D

Equate length measure of AD and CD

Say, if coordinate of D are

AD =

Squaring both sides,

AD2 =

Similarly, CD2 =

Since all sides of a square are equal, which means AD = CD

Value of y1 can be calculated as follows by using the value of x.

From step 2: each side of square = 2√2

CD2 =

8 =

Hence, D = (1, 4)

Step 4: Find coordinates of point B

From line segment, BOD

Coordinates of B can be calculated using coordinates of O; as follows:

Earlier, we had calculated O = (1, 2)

Say B =

For BD;

1 =

x2 = 1

And 2 =

=> y2 = 0

Therefore, the coordinates of required points are B = (1,0) and D = (1,4)

**Question 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?

**Answer 5:**

(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.

**Question 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 . Calculate the area of the ∆ ADE and compare it with area of ∆ ABC. (Recall Theorem 6.2 and Theorem 6.6)

**Answer 6:**

Given: The vertices of a ∆ ABC are A (4, 6), B (1, 5) and C (7, 2)

Point D and Point E divide AB and AC respectively in ratio 1 : 3.

Coordinates of D can be calculated as follows:

x = and y =

Here

Consider line segment AB which is divided by the point D at the ration 1:3.

x =

y =

Similarly, Coordinates of E can be calculated as follows:

x =

y = = 5

**Find Area of triangle:**

Using formula: Area of a triangle =

Area of triangle ∆ ABC can be calculated as follows:

= ½ [4(5 – 2) + 1( 2 – 6) + 7( 6 – 5)]

= ½ (12 – 4 + 7) = 15/2 sq unit

Area of ∆ ADE can be calculated as follows:

= ½ [4(23/4 – 5) + 13/4 (5 – 6) + 19/4 (6 – 23/4)]

= ½ (3 – 13/4 + 19/16)

= ½ ( 15/16 ) = 15/32 sq unit

Hence, ratio of area of triangle ADE to area of triangle ABC = 1 : 16.

**Question 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.

**Answer 7:**

(i) Coordinates of D can be calculated as follows:

Coordinates of D =

D

(ii) Coordinates of P can be calculated as follows:

Coordinates of P =

P

(iii) Coordinates of E can be calculated as follows:

Coordinates of E = = (5/2 , 3)

E(5/2 , 3)

Point Q and P would be coincident because medians of a triangle intersect each other at a common point called centroid. Coordinate of Q can be given as follows:

**Coordinates of Q** =

F is the mid- point of the side AB

Coordinates of F =

Point R divides the side CF in ratio 2:1**Coordinates of R** =

(iv) Coordinates of P, Q and R are same which shows that medians intersect each other at a common point, i.e. centroid

of the triangle.

(v) If A (x_{1}, y_{1}), B (x_{2}, y_{2}) and C (x_{3}, y_{3}) are the vertices of triangle ABC, the coordinates of centroid can be given as follows:

**Question 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.

**Answer 8:**

P id the mid-point of side AB,

Coordinate of P =

Similarly, Q, R and S are (As Q is mid-point of BC, R is midpoint of CD and S is midpoint of AD)

Coordinate of Q = (2, 4)

Coordinate of R = (5,

Coordinate of S = (2, -1)

Now,

Length of PQ = = =

Length of SP = = =

Length of QR = = =

Length of RS = = =

Length of PR (diagonal) = = 6

Length of QS (diagonal) = = 5

The above values show that, PQ = SP = QR = RS = , i.e. all sides are equal.

But PR ≠ QS i.e. diagonals are not of equal measure.

Hence, the given figure is a rhombus.

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