If $P_1$ and $P_2$ are perpendicular distances (in units) from point $(2,-1)$ to the pair of lines $2 x^2-5…
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
2
5
10
4
Solution
$\begin{array}{ll}
& \text { Given equation of pair of lines is } \\
& 2 x^2-5 x y+2 y^2=0 \\
\therefore \quad & 2 x^2-4 x y-x y+2 y^2=0 \\
\therefore \quad & 2 x(x-2 y)-y(x-2 y)=0 \\
\therefore \quad & (2 x-y)(x-2 y)=0
\end{array}$
$\therefore \quad$ separate equations of the lines are
$2 x-y=0 \text { and } x-2 y=0$
$\therefore \quad$ Perpendicular distances of the above lines from $(2,-1)$ are
$\begin{aligned} & P_1=\left|\frac{2(2)-(-1)}{\sqrt{(2)^2+(-1)^2}}\right|=\left|\frac{5}{\sqrt{5}}\right| \text { and } \\ & P_2=\left|\frac{2-2(-1)}{\sqrt{(1)^2+(-2)^2}}\right|=\left|\frac{4}{\sqrt{5}}\right|\end{aligned}$
$\therefore \quad P_1 P_2=\frac{5}{\sqrt{5}} \times \frac{4}{\sqrt{5}}=4$