First slide

Equation of line in Straight lines

Question

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

Moderate

A

$2 x - 3 y = 0$
B

$3 x - 2 y = 0$
C

$5 x - 7 y = 0$
D

$7 x - 5 y = 0$

Solution

Suppose that the coordinatres of the point on the line $x = 2 y$ is $(2 α, α)$ and suppose that the foot of the perpendciular from $P (2 α, α)$ on the line $x = y$ is $Q (β, β)$ . Let $R (h, k) b e t h e m i d p o i n t o f P Q .$
Slope of PQ $= \frac{k - α}{h - 2 α} = - 1$ , since $P Q$ is perpendicualr to the line $y = x$
$\Rightarrow k - α = - h + 2 α$
$\Rightarrow α = \frac{h + k}{3}$ …………. (1)
Also $2 h = 2 α + β$ and $2 k = α + β$
$\Rightarrow 2 h = α + 2 k$
$\Rightarrow α = 2 h - 2 k$ …………. (2)
From (1) and (2)
$\frac{h + k}{3} = 2 (h - k)$
$∴$ locus is $6 x - 6 y = x + y \Rightarrow 5 x = 7 y$

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