The locus of the mid-points of the
Question:

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

  1. (1) $2 x-3 y=0$

  2. (2) $5 x-7 y=0$

  3. (3) $3 x-2 y=0$

  4. (4) $7 x-5 y=0$


Correct Option: , 2

Solution:

Since, slope of $P Q=\frac{k-\alpha}{h-2 \alpha}=-1$

$\Rightarrow \quad k-\alpha=-h+2 \alpha$

$\Rightarrow \alpha=\frac{h+k}{3}$

Also, $2 h=2 \alpha+\beta$ and

$2 k=\alpha+\beta$

$\Rightarrow \quad 2 h=\alpha+2 k$

$\Rightarrow \quad \alpha=2 h-2 k$

From (i) and (ii), we have

$\frac{h+k}{3}=2(h-k)$

So, locus is $6 x-6 y=x+y$

$\Rightarrow 5 x=7 y \Rightarrow 5 x-7 y=0$

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