The product of lengths of the perpendiculars drawn from the point ( - 1 , 5 ) to the pair of lines 2 x 2 - x…

Given:

$2x^2-xy-3y^2+6x+y+4=0$

First, we will find the separate equations.

$2x^2-xy-3y^2=0$

$\Rightarrow 2x^2-xy+2xy-2xy-3y^2=0$

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

Now, $m \& l$ be the constants term in the above equations.

$x+y+m=0$

$2x-3y+l=0$

Then,

$2x^2-xy-3y^2+6x+y+4=\left(x+y+m\right)\left(2x-3y+l\right)$

On comparing the coefficients, we get

$lx+2mx=6x$

$\Rightarrow l+2m=6 ...\left(i\right)$

$ly-3my=y$

$\Rightarrow l-3m=1 ...\left(ii\right)$

Solving $\left(i\right) \& \left(ii\right)$, we get

$m=1 \& l=4$

So, we have the separate equations for lines are

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

$2x-3y+4=0 ...\left(iv\right)$

Now, we need to calculate $PM \& PN$.

$PM=\frac{\left|-1\times 1+5\times 1+1\right|}{\sqrt{1^2+1^2}}=\frac{5}{\sqrt{2}}$

$PN=\frac{\left|2\times \left(-1\right)+\left(-3\right)\times \left(5\right)+4\right|}{\sqrt{2^2+{\left(-3\right)}^2}}=\frac{13}{\sqrt{13}}$

Now,

$PM\times PN=\left(\frac{5}{\sqrt{2}}\times \frac{13}{\sqrt{13}}\right)$

                    $=\frac{65}{\sqrt{26}}$.