# KSEEB SSLC Class 10 Maths Solutions Chapter 7 Coordinate Geometry Ex 7.2

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 the given points be A(4, -1) and B(-2, -3) Let the points P and Q trisects AB.
i.e., AP = PQ = QB
i.e., P divides AB in the ratio of 1 : 2 and Q divides AB in the ratio of 2 : 1
Let the coordinates of P be (x, y) 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:
Let the given points are A(-3, 10) and B(6, -8).
Let the point P(-1, 6) divides AB in the ratio m1 : m2.
∴ Using the section formula, we have ⇒ -1(m1 + m2) = 6m1 – 3m2
and 6(m1 + m2) = – 8m1 + 10m2
⇒ -m1 – m2 – 6m1 + 3m2 = 0
and 6m1 + 6m2 + 8m1 – 10m2 = 0
⇒ -7m1 + 2m2 = 0 and 14m1 – 4m2 = 0 or 7m1 – 1m2 = 0
⇒ $$\frac{m_{1}}{m_{2}}=\frac{2}{7}$$ and $$\frac{m_{1}}{m_{2}}=\frac{2}{7}$$
⇒ 2m2 = 7m1 and 7m1 = 2m2
⇒ m1 : m2 = 2 : 7 and m1 : m2 = 2 : 7
Thus, the required ratio is 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:
Let the given points are A( 1, 2), B(4, y), C(x, 6) and D(3, 5) Since, the diagonals of a parallelogram bisect each other.
∴ The coordinates of P are : ∴ The required values of x and y are 6 and 3 respectively. 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:
Here, the given points are A(-2, 2) and B(2, 8)
Let P1, P2 and P3 divide AB in four equal parts.
<  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, 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