The product of the perpendicular distances from $(1,-1)$ to the pair of lines $x^2-4 x y+y^2=0$, is

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Aicharya · Accepted Answer

The given pair of straight lines, $ \begin{aligned} & x^2-4 x y+y^2=0 \ & \Rightarrow \quad x^2-4 x y+4 y^2=3 y^2 \ & \Rightarrow \quad(x-2 y)^2-(\sqrt{3} y)^2=0 \ & \Rightarrow \quad x-(2+\sqrt{3}) y=0 \ & 	ext { or } \quad x-(2-\sqrt{3}) y=0 \ & \end{aligned} $ So, perpendicular distances from point $(1,-1)$ to lines (i) $ d_1=\frac{|1+2+\sqrt{3}|}{\sqrt{1+(2+\sqrt{3})^2}}, d_2=\frac{|1+2-\sqrt{3}|}{\sqrt{1+(2-\sqrt{3})^2}} $ So, $ \begin{aligned} d_1 d_2 & =\frac{(3+\sqrt{3})}{\sqrt{8+4 \sqrt{3}}} 	imes \frac{(3-\sqrt{3})}{\sqrt{8-4 \sqrt{3}}} \ & =\frac{9-3}{\sqrt{64-48}}=\frac{6}{4}=\frac{3}{2} \end{aligned} $

The product of the perpendicular distances from $(1,-1)$ to the pair of lines $x^2-4 x y+y^2=0$, is

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