AP EAMCET · Maths · Pair of Lines
The product of the perpendicular distances from \((1,-1)\) to the pair of lines \(x^2-4 x y+y^2=0\), is
- A 1
- B \(\frac{2}{3}\)
- C \(\frac{3}{2}\)
- D 2
Answer & Solution
Correct Answer
(C) \(\frac{3}{2}\)
Step-by-step Solution
Detailed explanation
The given pair of straight lines, \[ \begin{aligned} & x^2-4 x y+y^2=0 \\ & \Rightarrow \quad x^2-4 x y+4 y^2=3 y^2 \\ & \Rightarrow \quad(x-2 y)^2-(\sqrt{3} y)^2=0 \\ & \Rightarrow \quad x-(2+\sqrt{3}) y=0 \\ & \text { or } \quad x-(2-\sqrt{3}) y=0 \\ & \end{aligned} \] So,…
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