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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 (x1, y1) and Q (x2, y2) 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.

x₁ = (1×(-2)+2×4)/3 = (-2+8)/3 = 6/3 = 2
y₁ = (1×(-3)+2×(-1))/(1+2) = (-3-2)/3 = -5/3

Therefore: P (x1, y1) = P(2, -5/3)

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

x₂ = (2×(-2)+1×4)/(2+1) = (-4+4)/3 = 0
y₂ = (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 = ( 6k-3)/(k+1)

–k – 1 = 6k -3
7k = 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 {x₁ (y₂ – y₃) + x₂ (y₃ – y₁) + x₃ (y₁ – y₂)}
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 (x₁, y₁), B (x₂, y₂) and C (x₃, y₃) 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 (x₁, y₁), B (x₂, y₂) and C (x₃, y₃) 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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