First slide

Equation of line in Straight lines

Question

The locus of midpoints of the perpendicular drawn from the points on the line $x = 3 y$ to the line x=y is

Moderate

A

$2 x - 3 y = 0$
B

$7 x - 5 y = 0$
C

$5 x - 3 y = 0$
D

$3 x - 5 y = 0$

Solution

The general point on the line $x = 3 y$ is $(3 h, h)$
The foot of the perpendicular of $(3 h, h)$ on the line $x = y$ is $(l, m)$
$\frac{l - 3 h}{1} = \frac{m - h}{- 1} = \frac{- (3 h - h)}{1^{2} + 1^{2}}$
Simplify
$\frac{l - 3 h}{1} = - h \Rightarrow l = 2 h$
And
$\frac{m - h}{- 1} = - h \Rightarrow m = 2 h$
Hence the foot of the perpendicular of $(3 h, h)$ on the line $x - y = 0$ is $(2 h, 2 h)$
Suppose that $P (x_{1,} y_{1})$ be any point on the locus.
The point $P (x_{1,} y_{1})$ is midpoint of $(3 h, h)$ and $(2 h, 2 h)$
It implies that $(x_{1}, y_{1}) = (\frac{5 h}{2}, \frac{3 h}{2})$
Eliminate h to get the locus of the point $P (x_{1,} y_{1})$
Therefore, the locus is $\frac{2 x}{5} = \frac{2 y}{3} \Rightarrow 3 x - 5 y = 0$

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