If \(P_1\) and \(P_2\) are perpendicular distances (in units) from point \((2,-1)\) to the pair of lines \(2 x^2-5 x y+2 y^2=0\), then the value of \(\mathrm{P}_1 \mathrm{P}_2\) is

Given equation of pair of lines is
\( 2 x^2-5 x y+2 y^2=0 \)
\( \therefore 2 x^2-4 x y-x y+2 y^2=0 \)
\( \therefore 2 x(x-2 y)-y(x-2 y)=0 \)
\( \therefore (2 x-y)(x-2 y)=0\)
\(\therefore\) separate equations of the lines are
\(2 x-y=0 \text { and } x-2 y=0\)
\(\therefore\) Perpendicular distances of the above lines from \((2,-1)\) are
\(\begin{aligned} & P_1=\left|\frac{2(2)-(-1)}{\sqrt{(2)^2+(-1)^2}}\right|=\left|\frac{5}{\sqrt{5}}\right| \text { and } \\ & P_2=\left|\frac{2-2(-1)}{\sqrt{(1)^2+(-2)^2}}\right|=\left|\frac{4}{\sqrt{5}}\right|\end{aligned}\)
\(\therefore \quad P_1 P_2=\frac{5}{\sqrt{5}} \times \frac{4}{\sqrt{5}}=4\)