Question

The line $(a+2b)x+(a-3b)y=a-b$ for different values of $a$ and $b$ passes through the fixed point

• A. $(\frac{3}{5},\frac{7}{5})$
• B. $(\frac{7}{2},\frac{5}{2})$
• C. $(\frac{6}{5},\frac{6}{5})$
• D. $(\frac{2}{5},\frac{3}{5})$

Answer 1

The correct option is D $(\frac{2}{5},\frac{3}{5})$

Equation of line is

$(a+2b)x+(a-3b)y=a-b$

$\Rightarrow (a+2b)x+(a-3b)y-a+b=0$

$\Rightarrow ax+2bx+ay-3by-a+b=0$

$\Rightarrow (x+y-1)a+(2x-3y+1)b=0$

Hence,

$x+y-1=0......\left(1\right)$

$2x-3y+1=0......\left(2\right)$

On solving ( 1) and ( 2) to and we get,

$x=\frac{2}{5},y=\frac{3}{5}$

The points $(\frac{2}{5},\frac{3}{5})$

Hence, this is the answer.