Q. 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) $\left( \frac{2}{5},\frac{3}{5} \right)$
Explanation
(d)
Given equation can be rewritten as
$a(x + y - 1) + b(2x - 3y + 1) = 0$
This is the form of intersection of two lines.
$\therefore\ x + y - 1 = 0$ …(i)
$and\ \ 2x - 3y + 1 = 0$ ….(ii)
On solving Eqs. (i) and (ii), we get
$x = \frac{2}{5}\ and\ y = \frac{3}{5}$
Hence, coordinates of required point are $\left( \frac{2}{5},\frac{3}{5} \right)$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.