Question

Find the ratio in which the line seg ment joining (-2, -3) and (5, 6) is divided by y-axis. [CBSE 2012]​

Answer 1

Step-by-step explanation:

Given :-

The points (-2, -3) and (5, 6)

To find :-

Find the ratio in which the line seg ment joining (-2, -3) and (5, 6) is divided by y-axis ?

Solution :-

Given points are (-2, -3) and (5, 6)

We know that

The equation of y-axis is x = 0

The required point = (0,y)

Let the required ratio be m1:m2

Let (x1, y1) = (-2,-3) =>x1 = -2 and y1 = -3

Let (x2, y2) = (5,6) => x2 = 5 and y2 = 6

We know that

The coordinates of the point P(x,y) which divides the linesegment joining the points (x1, y1) and

(x2, y2) in the ratio m1:m2 is

({m1x2+m2x1}/(m1+m2) , {m1y2+m2y1}/(m1+m2))

On substituting these values in the given formula then

=> (0,y) =

({m1×5+m2×-2}/(m1+m2) , { m1×6+m2×-3}/(m1+m2))

=> (0,y) = ({5m1-2m2}/(m1+m2),{6m1-3m2}/(m1+m2))

On comparing both sides then

=> {5m1-2m2}/(m1+m2) = 0

=> 5m1 -2m2 = 0(m1+m2)

=> 5m1-2m2 = 0

=> 5m1 = 2m2

=> m1/m2 = 2/5

=> m1:m2 = 2:5

and

y = {m1×6+m2×-3}/(m1+m2)

=> y = {(6)(2)+(5)(-3)}/(2+5)

=> y = (12-15)/7

=> y = -3/7

The required point = (0,-3/7)

Answer:-

The required ratio for the given problem is 2:5

Used formulae:-

→ The coordinates of the point P(x,y) divides the linesegment joining the points (x1,y1) and (x2, y2) in the ratio m1:m2 is

({m1x2+m2x1}/(m1+m2) , {m1y2+m2y1}/(m1+m2))

→ The equation of y-axis is x = 0.