KSEEB SSLC Class 10 Maths Solutions Chapter 7 Coordinate Geometry Ex 7.2 are part of KSEEB SSLC Class 10 Maths Solutions. Here we have given Karnataka SSLC Class 10 Maths Solutions Chapter 7 Coordinate Geometry Exercise 7.2.

## Karnataka SSLC 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) is the required point.
P(x, y) divides the points A(-1, 7),
B (4, -3) in the ratio m1 : m2 = 2 : 3.
As per Section Formula, coordinates of P(x, y) are

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

Solution:
Let A(x1, y1) = A(4,-1)
B(x2, y2) = B(-2, -3).
P and Q are the points of trisection of AB.
AP = PQ = QB
i) P divides AB in the ratio m1 : m2 :: 1 : 2.
∴ As per Section Formula, Coordinates of P are :

ii) Q divides AB in the ratio 2 : 1.
∴ As per Section formula,

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 the following figure.

Niharika runs $$\frac{1}{4}$$ th the distance of AD on then 2nd line and posts a green flag
Preet runs $$\frac{1}{5}$$ th the distance AD on the 5 eighthline 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:
Origin, A(0, 0)
AB along x – axis
In the parallel line AD in 2nd row, Rashmi sees green flag at the distance of $$\frac{1}{4}$$ .
∴ AD = $$\frac{1}{4}$$ × 100 = 25 m.
∴ Coordinates of green flag are G(2, 25)
Similarly we can find out coordinates of Red flag as R(8, 20).
∴ Distance between G and R is

It means blue flag is in 5th line and parallel to AD at the distance of 22.5 m.

Question 4.
Find the ratio in which the line segment joining the points (-3, 10) and (6, -8) is divided by (-1, 6).
Solution:
We have to check whether the line segment joining the points (-3, 10) and (6, -8) is divided by (-1, 6).
Let P(- 1, 6) divides AB in the ratio m1 : m2.
By using Section formula,

∴ P(-1, 6) point divides the line segment in the ratio 2 : 7.

Question 5.
Find the ratio in whcih 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 line segment joining A and B points divides x-axis in the ratio k : 1.
If coordinates of y, P(x, 0) compared,
$$\frac{k \times(5)+1 \times(-5)}{k+1}=0$$
5k – 5 = 0
∴ k = 1
∴ Ratio will be k : 1 = 1 : 1
∴ Point P(x, 0) is the mid-point of line segment AB.
∴ x = $$\frac{1-4}{2}=\frac{-3}{2}$$
∴ Dividing point, P($$-\frac{3}{2}$$, 1)

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.

Solution:
AC and BD are the diagonals of a parallelogram ABCD and diagonals bisect at O’.
∴ Coordinate of the midpoint of AC = Coordinate of midpoint BD.
As per Mid-point formula,

∴ y = 3
∴ x = 6, y = 3.

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

Solution:
AB is the diameter.
‘O’ is centre of the circle, bisects AB
Let coordinates of A are (x, y)
As per the Mid-point formula,

∴ The coordinates of point A are (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 coordinates of P are (x, y).
AP = $$\frac{3}{7}$$ AB AP : AB = 3 : 7
As per Section formula,
AP : AB = 3 : 7.
∴ AP + PB = AB
3 + PB = 7
∴ PB = = 7 – 3 = 4
∴ Let AP : PB = 3 : 4 = m1 : m2.
Coordinates of P are

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 A (-2, 2) = (x1, y1).
B (2, 8) = (x4, y4).
P, Q, R points divide equally the side AB.
Here, we have AP = PQ = QR = RB.
∴ AP : PB = 1 : 3
AQ : QB = 2 : 2 = 1 : 1
∴ AR : RB = 3 : 1.
As per Section formula
i) P divides AB in the ratio 1 : 3

ii) Q divides AB in the ratio 2 : 2 means it is mid-point of AB.
∴ As per the Mid-point formula,

Q(x2, y2) = (0, 5)

iii) R divides AB in the ratio 3 : 1
m1 = 3, m2 = 1

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 = $$\frac{1}{2}$$ (product of its diagonals]

Solution:
ABCD is a rhombus.
AC and BD are the diagonals and intersects at ‘O’ and bisects perpendicularly.
∴ Let Diagonal BD = d1 Diagonal AC = d2,

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