**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 m_{1} : m_{2} = 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(x_{1}, y_{1}) = A(4,-1)

B(x_{2}, y_{2}) = B(-2, -3).

P and Q are the points of trisection of AB.

AP = PQ = QB

i) P divides AB in the ratio m_{1} : m_{2} :: 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

AD along y – axis

In the parallel line AD in 2^{nd} 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 m_{1} : m_{2}.

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 = m_{1 }: m_{2}.

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) = (x_{1}, y_{1}).

B (2, 8) = (x_{4}, y_{4}).

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(x_{2}, y_{2}) = (0, 5)

iii) R divides AB in the ratio 3 : 1

m_{1} = 3, m_{2} = 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 = d_{1} Diagonal AC = d_{2},

We hope the given KSEEB SSLC Class 10 Maths Solutions Chapter 7 Coordinate Geometry Ex 7.2 will help you. If you have any query regarding Karnataka SSLC Class 10 Maths Solutions Chapter 7 Coordinate Geometry Exercise 7.2, drop a comment below and we will get back to you at the earliest