MHT CET · Maths · Pair of Lines
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
- A \(\frac{9}{\sqrt{5}}\) units
- B \(\frac{1}{\sqrt{5}}\) units
- C 4 units
- D 9 units
Answer & Solution
Correct Answer
(C) 4 units
Step-by-step Solution
Detailed explanation
\(2 x^2-5 x y+2 y^2=0 \)
\( \therefore 2 x^2-4 x y-x y+2 y^2=0 \Rightarrow 2 x(x-2 y)-\)\(y(x-2 y)=0 \)
\( \therefore(2 x-y)(x-2 y)=0\)
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 \(\frac{5}{\sqrt{5}} \times \frac{4}{\sqrt{5}}=5\)
\( \therefore 2 x^2-4 x y-x y+2 y^2=0 \Rightarrow 2 x(x-2 y)-\)\(y(x-2 y)=0 \)
\( \therefore(2 x-y)(x-2 y)=0\)
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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