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The product of the perpendicular distances from $(2,-1)$ to the pair of lines $2 x^2-5 x y+2 y^2=0$ is
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Verified Answer
The correct answer is:
4 units
$$
\begin{aligned}
& 2 x^2-5 x y+2 y^2=0 \\
& \therefore 2 x^2-4 x y-x y+2 y^2=0 \Rightarrow 2 x(x-2 y)-y(x-2 y)=0 \\
& \therefore(2 x-y)(x-2 y)=0
\end{aligned}
$$
Thus lines are $2 x-y=0$ and $x-2 y=0$
Distance of point $(2,-1)$ from these two lines are respectively $\left|\frac{(2)(2)+(-1)(1)}{\sqrt{4+1}}\right|$ and $\left|\frac{(1)(2)+(-1)(-2)}{\sqrt{1+4}}\right|$ i.e. $\frac{5}{\sqrt{5}}$ and $\frac{4}{\sqrt{5}}$ Hence required answer is $\frac{5}{\sqrt{5}} \times \frac{4}{\sqrt{5}}=5$
\begin{aligned}
& 2 x^2-5 x y+2 y^2=0 \\
& \therefore 2 x^2-4 x y-x y+2 y^2=0 \Rightarrow 2 x(x-2 y)-y(x-2 y)=0 \\
& \therefore(2 x-y)(x-2 y)=0
\end{aligned}
$$
Thus lines are $2 x-y=0$ and $x-2 y=0$
Distance of point $(2,-1)$ from these two lines are respectively $\left|\frac{(2)(2)+(-1)(1)}{\sqrt{4+1}}\right|$ and $\left|\frac{(1)(2)+(-1)(-2)}{\sqrt{1+4}}\right|$ i.e. $\frac{5}{\sqrt{5}}$ and $\frac{4}{\sqrt{5}}$ Hence required answer is $\frac{5}{\sqrt{5}} \times \frac{4}{\sqrt{5}}=5$
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