The lines $(a+2 b) x+(a-3 b) y=a-b$ for different values of $a$ and $b$ pass through the fixed point whose…
The lines $(a+2 b) x+(a-3 b) y=a-b$ for different values of $a$ and $b$ pass through the fixed point whose coordinates are
$\left(\frac{2}{5}, \frac{2}{5}\right)$
$\left(\frac{3}{5}, \frac{3}{5}\right)$
$\left(\frac{2}{5}, \frac{3}{5}\right)$
$\left(\frac{3}{5}, \frac{2}{5}\right)$
Solution
Given, equation of line is
$
\begin{array}{rlrl}
& & (a+2 b) x+(a-3 b) y & =a-b \\
\Rightarrow & a x+2 b x+a y-3 b y & =a-b \\
\Rightarrow & a(x+y-1)+b(2 x-3 y+1) & =0
\end{array}
$
Equating the coefficient, we obtain
$
\begin{array}{r}
x+y-1=0 \\
2 x-3 y+1=0
\end{array}
$
Solve Eqs. (i) and (ii) for $x, y$, we obtain
$
x=2 / 5, y=3 / 5
$
$\therefore$ Desired point of coordinate is $\left(\frac{2}{5}, \frac{3}{5}\right)$