A straight line passes through a fixed point | vedclass

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(A) Let the foot of the perpendicular from the origin $(0, 0)$ to the line be $(x, y)$.Since the line passes through $(h, k)$ and $(x, y)$,its slope i

Vedclass · Accepted Answer

$x^2 + y^2 - hx - ky = 0$

Vedclass · Answer

(A) Let the foot of the perpendicular from the origin $(0, 0)$ to the line be $(x, y)$.Since the line passes through $(h, k)$ and $(x, y)$,its slope is $m = \frac{y - k}{x - h}$.The slope of the perpendicular line segment from the origin to $(x, y)$ is $m' = \frac{y - 0}{x - 0} = \frac{y}{x}$.Since the lines are perpendicular,the product of their slopes is $-1$:$\left(\frac{y - k}{x - h}ight) 	imes \left(\frac{y}{x}ight) = -1$$y(y - k) = -x(x - h)$$y^2 - ky = -x^2 + hx$$x^2 + y^2 - hx - ky = 0$Thus,the required locus is $x^2 + y^2 - hx - ky = 0$.

$A$ straight line passes through a fixed point $(h, k)$. The locus of the foot of the perpendicular drawn from the origin to this line is:

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