2. Find the point A (x, y) when C divides AB in the ratio 2:3. Points B and C are given as
(-5, 8) and (3,6).
Answers
Given:
C divides AB in the ratio 2:3
Points B and C are (-5, 8) and (3,6).
To find:
The coordinates of point A
Solution:
Let the coordinate of A (a,b)
We will use the section formula to get the coordinate of A which is given as:
(x,y) = (mx2+nx1/(m+n), my2+ny1/(m+n))
according to the question:
- (3,6) = (2.(-5)+3a/(3+2), 2.8+3.b/(3+2))
- (3,6) = (-10+3a/5, 16+3.b/5)
on comparing we get:
3 = -10+3a/5
- -10+3a = 15
- 3a = 15+10
- 3a = 25
- a = 25/3
6 = 16+3b/5
- 16+3b = 30
- 3b = 30-16
- 3b = 14
- b = 14/3
So the coordiantes of A are (25/3, 14/3)
Given : C divides AB in the ratio 2:3.Points B and C are given as (-5,8) and (3,6)
To Find : point A(x,y)
Solution:
A (x,y)
B (-5,8)
C (3,6)
C divides AB in 2:3 ratio
=> Cx = (2Bx + 3Ax)/(2 + 3)
Cy = (2By + 3Ay)/(2 + 3)
Cx = (2Bx + 3Ax)/(2 + 3)
=> 3 = ( 2 *(-5) + 3x) /5
=>15 = -10 + 3x
=> 3x = 25
=> x = 25/3
Cy = (2By + 3Ay)/(2 + 3)
=> 6 = ( 2 *8 + 3y) /5
=>30 = 16 + 3y
=> 3y = 14
=> y = 14/3
point A(x,y) = ( 25/3 , 14/3)
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