AP EAMCET · Maths · Straight Lines
If \(p_1\) and \(p_2\) denote the lengths of perpendiculars from \((2,3)\) onto the lines given by \(15 x^2+31 x y+14 y^2=0\), and if \(p_1>p_2\), then \(p_1^2+\frac{1}{74}-p_2^2+\frac{1}{13}\) is equal to
- A -2
- B 0
- C 2
- D 1
Answer & Solution
Correct Answer
(C) 2
Step-by-step Solution
Detailed explanation
Given equation, \(15 x^2+31 x y+14 y^2=0\) \[ \Rightarrow \quad \begin{aligned} 15 x^2+10 x y+21 x y+14 y^2 & =0 \\ 5 x(3 x+2 y)+7 y(3 x+2 y) & =0 \\ (3 x+2 y)(5 x+7 y) & =0 \end{aligned} \] \(\therefore\) Two lines have following equation…
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