The locus of the mid-points of the perpendiculars drawn from points on the line x = 2 y , to the line x = y …

1. $2x-3y=0$
2. $5x-7y=0$
3. $3x-2y=0$
4. $7x-5y=0$

Solution

Slope of $\mathrm{P}\mathrm{Q}=\frac{\mathrm{k}-\mathrm{\alpha }}{\mathrm{h}-2\mathrm{\alpha }}=-1$
$k-\alpha =-h+2\alpha$ $\Rightarrow \alpha =\frac{h+k}{3} \dots \left(1\right)$
Also, $2h=2\alpha +\beta$ and $2k=\alpha +\beta$
$2h=\alpha +2k$$\Rightarrow \alpha =2h-2k \dots \left(2\right)$
From $\left(1\right) \& \left(2\right)$,
$\frac{h+k}{3}=2\left(h-k\right)$
So, locus is $6x-6y=x+y\Rightarrow 5x=7y$.

Asked in: JEE Main 2020 (07 Jan Shift 2)

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