Question

Find the coordinates of the point which divides the line segment joining the points (a+b,a-b) and (a-b,a+b) in the ratio 3:2 internally

Answer 1

Answer:

Step-by-step explanation:

Given points:

(a+b,a-b) and (a-b,a+b)

In the ratio 3:2

=3(a-b)+2(a+b)/3+2,3(a+b)+2(a-b)/3+2

=3a-3b+2a+2b/5,3a+3b+2a-2b/5

=5a-b/5,5a+b/5

Answer 2

Answer:

$(\frac{5a-b}{5},\frac{5a+b}{5} )$

Step-by-step explanation:

Given: Two points $(a+b,a-b)$ and $(a-b,a+b)$, internal division is in the ratio $3:2$.

To find coordinate of point.

If line segment joining points $(x_{1},y_{1} )$ & $(x_{2},y_{2} )$ is divided in the ratio $m:n$ then the coordinate of point is given by $(\frac{mx_{2}+nx_{1} }{m+n} ,\frac{my_{2}+ny_{1} }{m+n} )$.

Here, $(x_{1},y_{1} )=(a+b,a-b), (x_{2},y_{2} )=(a-b,a+b)$ and $m:n=3:2$.

$[\frac{3(a-b)+2(a+b)}{3+2}, \frac{3(a+b)+2(a-b)}{2+3} ]$

$(\frac{3a-3b+2a+2b}{5},\frac{3a+3b+2a-2b}{5} )\\(\frac{5a-b}{5},\frac{5a+b}{5} )$

So, the coordinate of point is $(\frac{5a-b}{5},\frac{5a+b}{5} )$.

#SPJ2