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