JEE Mains · Maths · STD 11 - 9. straight line
The locus of the mid-points of the perpendiculars drawn from points on the line, \(\mathrm{x}=2 \mathrm{y}\) to the line \(\mathrm{x}=\mathrm{y}\) is
- A \(2 x-3 y=0\)
- B \(7 x-5 y=0\)
- C \(5 x-7 y=0\)
- D \(3 x-2 y=0\)
Answer & Solution
Correct Answer
(C) \(5 x-7 y=0\)
Step-by-step Solution
Detailed explanation
\(\frac{\alpha-\beta}{2 \alpha-\beta}=-1\) \(3 \alpha=2 \beta\) \(\mathrm{h}=\frac{2 \alpha+\beta}{2}\) \(2 \mathrm{h}=\frac{7 \alpha}{2}\) \(\mathrm{k}=\frac{\alpha+\beta}{2}\) \(2 \mathrm{k}=\frac{5 \alpha}{2}\) \(\frac{\mathrm{h}}{\mathrm{k}}=\frac{7}{5}\) \(5 x=7 y\)
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