A) \[\frac{5}{12}\]
B) \[\frac{12}{5}\]
C) \[\frac{6}{5}\]
D) \[\frac{5}{6}\]
E) \[\frac{1}{5}\]
Correct Answer: B
Solution :
\[2{{x}^{2}}-5xy+2{{y}^{2}}+3x-3y+1=0\] \[\Rightarrow \]\[(x-2y+1)(2x-y+1)=0\] \[\therefore \]Two equations are\[x-2y+1=0\]and\[2x-y+1=0\]. Length of perpendiculars from\[(-1,2)\]are \[{{P}_{1}}=\left| \frac{-1-4+1}{\sqrt{1+4}} \right|=\frac{4}{\sqrt{5}}\] and \[{{P}_{2}}=\left| \frac{-2-2+1}{\sqrt{5}} \right|=\frac{3}{\sqrt{5}}\] \[\therefore \]Product\[={{p}_{1}}.{{p}_{2}}=\frac{12}{5}\]
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