RBSE Solutions for Class 10 Maths Chapter 7 Coordinate Geometry Ex 7.2

Rajasthan Board RBSE Solutions for Class 10 Maths Chapter 7 Coordinate Geometry Ex 7.2 Textbook Exercise Questions and Answers.

RBSE Class 10 Maths Solutions Chapter 7 Coordinate Geometry Exercise 7.2

Question 1.
Find the coordinates of the point which divides the join of (- 1, 7) and (4, – 3) in the ratio 2 : 3.
Solution:
Let P(x, y) be the required point which divides the line segment joining the given points A(- 1, 7) and B(4, – 3) in the ratio 2 : 3.

So required point = P(1, 3)

Question 2.
Find the coordinates of the points of trisection of the line segment joining (4, -1) and (- 2, – 3).
Solution:
Let P(x₁, y₁) and Q(x₂, y₂) be the required points which trisect the line segment joining the points A(4, – 1) and (- 2, – 3) i.e. P(x₁, y₁) divides AB in the ratio 1 : 2 and Q(x₂, y₂) divides AB in the ratio 2 : 1.

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 1m from each other along AD, as shown in Fig. Niharika

runs \(\frac{1}{4}\)th the distance AD on the 2nd line nd posts a green flag. Preet runs \(\frac{1}{5}\)th 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:
In the given figure we assume A as the origin. Taking AB as x-axis and AD as y-axis
Position of the green flag
= Distance covered by Niharika
[Niharika runs a distance equal to \(\frac{1}{4}\) part of AB on the 2nd line ]
= \(\frac{1}{4}\) × 100 = 25 m
∴ Coordinates of the green flag = (2, 25)
Now, position of the red flag
= Distance covered by Preet
[Preet runs a distance equal to \(\frac{1}{5}\) part of AB on eighth line]
= \(\frac{1}{5}\) × 100 = 20 m
∴ Coordinates of the red flag = (8, 20)
∴ Distance between the green and red flags
= \(\sqrt{(8-2)^{2}+(20-25)^{2}}\)
= \(\sqrt{36+25}\)
= \(\sqrt{61}\) m
Position of the blue flag
= Half of the line segment joining the green flag and the red flag
= (\(\frac{2+8}{2}\), \(\frac{25+20}{2}\))
= (5, 22.5)
Hence the blue flag is in the 5th line and is at a distance of 22.5 m towards AD.

Question 4.
Find the ratio in which the line segment joining the points (- 3, 10) and (6, – 8) is divided by (-1, 6)
Solution:
Let the point P(- 1, 6) divides the line segment joining the points A(- 3, 10) and B(6, – 8) in the ratio K : 1.

K : 1
or – K – 1 = 6K – 3
or -K – 6K = -3 + 1
or -7K = -2
or K = \(\frac{2}{7}\)
∴ K : 1 = \(\frac{2}{75}\) : 1 = 2 : 7
So required ratio = 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.
Solution:
Let the required point on the x-axis be P(x, 0) which divides the line segment joining A(1, – 5) and B(- 4, 5) in the ratio of K : 1.

or 5K – 5 = 0
or 5K = 5
or K = \(\frac{5}{5}\) = 1
∴ required ratio K : 1 = 1 : 1
Now, x-coordinate of P,
x = \(\frac{-4 \times K+1 \times 1}{K+1}\)
Substituting K = 1,
x = \(\frac{-4 \times 1+1 \times 1}{1+1}\) = \(\frac{-4+1}{2}\)
⇒ x = \(-\frac{3}{2}\)
Hence requited point = (\(-\frac{3}{2}\), 0)

Question 6.
If (1, 2), (4, y), (x, 6) and (3, 5) are the vertices of parallelogram taken in order, find x and y.
Solution:
Let the vertices of the parallelogram ABCD be A(l, 2), B(4, y), C(x, 6) and D(3, 5).
But the diagonal of a parallelogram bisect each other.
First Case : When E is the mid-point of A(1, 2) and C(x, 6)

Second Case—When E is the mid-point of B(4, y) and D(3, 5)
∴ Coordinates of E = (\(\frac{3+4}{2}\), \(\frac{5+y}{2}\))
⇒ E = (\(\frac{7}{2}\), \(\frac{5+y}{2}\)) …(ii)
But in (i) and (ii), the values of E are the same. Therefore comparing the coordinates.
\(\frac{x+1}{2}\) = \(\frac{7}{2}\) and 4 = \(\frac{5+y}{2}\)
or x + 1 = 7 or 8 = 5 + y
or x = 6 or y = 3
So the values of x and y are 6 and 3 respectively.

Question 7.
Find the coordinates of a point A, where AB is the diameter of a circle whose centre is (2, – 3) and (1, 4).
Solution:
Let the coordinates of A be (x, y).
But, the mid-point of the ends of a diameter is the centre.
∴ O is the mid-point of A(x, y) and B is (1, 4).
∴ (\(\frac{x+1}{2}\), \(\frac{y+4}{2}\)) = (2, -3)

Comparing
\(\frac{x+1}{2}\) = 2 and \(\frac{y+4}{2}\) = -3
or x + 1 = 4 or y + 4 = -6
or x = 3 or y = – 10
Hence the required point A = (3, – 10)

Question 8.
If A and B are (- 2, – 2) and (2, – 4), respectively, find the coordinates of P such that AP = \(\frac{3}{7}\) AB and P lies on the line segment AB.
Solution:
Let the required point be P(x, y) and according to the question
AP = \(\frac{3}{7}\) AB
But PB = AB – AP
= AB – \(\frac{3}{7}\) AB = (\(\frac{7-3}{7}\))AB
= \(\frac{4}{7}\) AB
∴ \(\frac{\mathrm{AP}}{\mathrm{PB}}\) = \(\frac{\frac{3}{7} \mathrm{AB}}{\frac{4}{7} \mathrm{AB}}\) = \(\frac{3}{4}\)
∴ P(x, y) divides the given points A and B in the ratio 3 : 4.
From Internal division formula.

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.
Solution:
Let C, D and E be the required points which divide the line segment joining the points A(- 2, 2) and B(2, 8) into four equal parts,

Then D in the mid-point of A and B; C is the mid-point of A and D; E is the mid¬point of D and B, so that
AC = CD = DE = EB
Now, the coordinates of the mid-point of A and B (i.e., coordinates of D)
= (\(\frac{-2+2}{2}\), \(\frac{2+8}{2}\))
= (0, 5)
The coordinates of the mid-point of A and D (i.e., coordinates of C)
= (\(\frac{-2+0}{2}\), \(\frac{2+5}{2}\)) = (-1, \(\frac{7}{2}\))
The coordinates of the mid-point of D and B (i.e., coordinates of E)
= (\(\frac{2+0}{2}\), \(\frac{8+5}{2}\)) = (1, \(\frac{13}{2}\))
Therefore, C, D and E are the required points.
Hence the coordinates of the points dividing AB into four equal parts are :
(0, 5), (-1, \(\frac{7}{2}\)) and (1, \(\frac{13}{2}\))

Question 10.
Find the area of rhombus if its vertices are (3, 0), (4, 5), (- 1, 4) and (- 2,- 1) taken in order. [Hint : Area of a
rhombus = \(\frac{1}{2}\)(product of its diagonals)]
Solution:
Let the vertices of the rhombus ABCD be A(3, 0); B(4, 5); C(- 1, 4) and D(- 2, – 1).


= 24 square units
Therefore, the area of the rhombus is 24 square units.