Locus of mid-points of perpendiculars drawn from points on line, mathrm{x}=2 mathrm{y} to line mathr

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9.Straight Line

hard

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

$3 x-2 y=0$

(JEE MAIN-2020)

Solution

$\frac{\alpha-\beta}{2 \alpha-\beta}=-1$

$3 \alpha=2 \beta$

$\mathrm{h}=\frac{2 \alpha+\beta}{2}$

$2 \mathrm{h}=\frac{7 \alpha}{2}$

$\mathrm{k}=\frac{\alpha+\beta}{2}$

$2 \mathrm{k}=\frac{5 \alpha}{2}$

$\frac{\mathrm{h}}{\mathrm{k}}=\frac{7}{5}$

$5 x=7 y$

Standard 11

Mathematics

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