Question

The lines $ (a+2b)x+(a-3b)y=a-b $ for different values of $a$ and $b$ pass through the fixed point whose coordinates are

• A. $ \left( \frac{2}{5},\frac{2}{5} \right) $
• B. $ \left( \frac{3}{5},\frac{3}{5} \right) $
• C. $ \left( \frac{1}{5},\frac{1}{5} \right) $
• D. $ \left( \frac{3}{5},\frac{2}{5} \right) $
• E. $ \left( \frac{2}{5},\frac{3}{5} \right) $

Answer 1

Answer: (E)

Given equation is $ (a+2b)\text{ }x+(a-3b)\text{ }y=a-b $
It can be rewritten as
$ a(x+y-1)+b(2x-3y+1)=0 $
This is the form of intersection of two lines.
$ \therefore $ $ x+y-1=0 $ and $ 2x-3y+1=0 $
On solving, we get
$ x=\frac{2}{5} $ and $ y=\frac{3}{5} $