AP EAMCET · Maths · Pair of Lines
Let \(L\) be the line joining the origin to the point of intersection of the lines represented by \(2 x^2-3 x y\) \(-2 y^2+10 x+5 y=0\). If \(L\) is perpendicular to the line \(k x+y+3=0\), then \(k\) is equal to
- A \(\frac {1}{2}\)
- B \(\frac {-1}{2}\)
- C \(-1\)
- D \(\frac {1}{3}\)
Answer & Solution
Correct Answer
(B) \(\frac {-1}{2}\)
Step-by-step Solution
Detailed explanation
Given that, \(\begin{aligned} & 2 x^2-3 x y-2 y^2+10 x+5 y=0 \\ & (2 x+y)(x-2 y+5)=0 \\ & 2 x+y=0 \text { and } x-2 y+5=0 \end{aligned}\) Now, equation of line passing through origin is \(2 x^2+y=0 \quad \Rightarrow m_1=-2\) Since, this line is perpendicular to the line…
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