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Pair of Lines

A straight li...

Question

A straight line passes through the point of intersection of the lines x+2y=1 and y=2. The locus of the foot of perpendicular on it drawn from origin is:

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Solution

$\begin{matrix} x + 2 y = 1 y = 2 P o i n t o f i n t e r sec t i o n = (- 3, 2) L e t t h e l o c u s o f f o o t o f p e r p e n d i c u l a r = (h, k) E q u a t i o n o f l i n e (y - k) = - \frac{h}{k} (x - h) k y - k^{2} = - h x + h^{2} h x + k y = (h^{2} + k^{2}) - 3 h + 2 k = h^{2} + k^{2} h^{2} + k^{2} + 3 h - 2 k = 0 L o c u s o f f o o t o f p e r p e n d i c u l a r = x^{2} + y^{2} + 3 x - 2 y = 0 \end{matrix}$

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