If a variable straight line passing through the point of intersection of the lines $x-2 y+3=0$ and $2…
If a variable straight line passing through the point of intersection of the lines $x-2 y+3=0$ and $2 x-y-1=0$ intersects the $\mathrm{X}, \mathrm{Y}$-axes at A and B respectively, then the equation of the locus of a point which divides the segment AB in the ratio $-2: 3$ is
$14 x^2+3 x y-15 y^2=0$
$x y=14 x+15 y$
$x^2+x y-y^2=0$
$14 x+3 x y-15 y=0$
Solution
Since, equation of a line passing through the two given lines is $x-2 y+3+K(2 x-y-1)=0$ $\Rightarrow(2 K+1) x+(-K-2) y-K+3=0$
$\Rightarrow \frac{(2 K+1)}{(K-3)} x+\frac{(-K-2)}{(K-3)} y=1$
So, $A\left(\frac{K-3}{2 K+1}, 0\right), B\left(0, \frac{3-K}{K+2}\right)$
Now, $(x, y)=\left(\frac{0+\frac{3(K-3)}{2 K+1}}{1}, \frac{\frac{2(K-3)}{K+2}}{1}\right)$
$\begin{aligned}
\Rightarrow x & =\frac{3 K-9}{2 K+1} \Rightarrow K=\frac{x+9}{3-2 x} \\
y & =\frac{2(K-3)}{K+2} \Rightarrow K=\frac{6+2 y}{2-y}
\end{aligned}$
So, $\frac{x+9}{3-2 x}=\frac{6+2 y}{2-y} \Rightarrow 14 x+3 x y-15 y=0$.