Question

Find the coordinates of the point which divides the line segment joining the poi
(a+b, a - b) and (a - b, a+b) in the ratio 3 : 2 internally.
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Answer 1

$\huge\rm\underline\purple{\underline{Solution:-}}$

Let the coordinates of the point be P(x,y)

The point P which divides the line segment joining (x₁,y₁) and (x₂,y₂) in the ratio m:n internally is given by ,

$\large\rm\bold{\boxed{P\:=\:(\frac{mx_2+nx_1}{m+n},\frac{my_2+ny_1}{m+n})}}$

Given Ratio = 3:2 (m = 3 , n = 2 )

The coordinates of points = (a+b,a-b) , ( a-b,a+b)

$\large\rm{\implies{(x,y)\:=\:(\frac{3(a-b)+2(a+b)}{5},\frac{3(a+b)+2(a-b)}{5})}}$

$\large\rm{\implies{(x,y)\:=\:(\frac{3a-3b+2a+2b}{5},\frac{3a+3b+2a-2b}{5})}}$

$\large\rm{\implies{(x,y)\:=\:(\frac{5a-b}{5},\frac{5a+b}{5})}}$

$\large\rm{\therefore{The\:point\:is\:(\frac{(5a-b)}{5},\frac{(5a+b)}{5})}}$