The locus of the mid-points of the perpendicu | vedclass

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(C) Let a point on the line $x=2y$ be $P(2\alpha, \alpha)$.Let the perpendicular from $P$ to the line $x-y=0$ meet it at $Q(\beta, \beta)$.The slope o

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$5x-7y=0$

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(C) Let a point on the line $x=2y$ be $P(2\alpha, \alpha)$.Let the perpendicular from $P$ to the line $x-y=0$ meet it at $Q(\beta, \beta)$.The slope of $PQ$ is $\frac{\alpha-\beta}{2\alpha-\beta}$.Since $PQ$ is perpendicular to $x-y=0$ (slope $1$),the slope of $PQ$ must be $-1$.So,$\frac{\alpha-\beta}{2\alpha-\beta} = -1 \implies \alpha-\beta = -2\alpha+\beta \implies 3\alpha = 2\beta \implies \beta = \frac{3\alpha}{2}$.Let $(h, k)$ be the mid-point of $PQ$.$h = \frac{2\alpha+\beta}{2} = \frac{2\alpha + \frac{3\alpha}{2}}{2} = \frac{7\alpha}{4}$.$k = \frac{\alpha+\beta}{2} = \frac{\alpha + \frac{3\alpha}{2}}{2} = \frac{5\alpha}{4}$.Now,$\frac{h}{k} = \frac{7\alpha/4}{5\alpha/4} = \frac{7}{5}$.$5h = 7k \implies 5x-7y=0$.

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

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