Question

A straight line passes through a fixed point (h,k).The locus of the foot of the perpendicular on it drawn from the origin

Answer 1

Let P(h, k) be the given point, let Q(x, y) be the foot of the perpendicular, and let O be the origin. The line can have any direction.  

∠PQO = 90°  

Point Q lies on the circle having diameter OP.  

The locus of point Q:  

(x - 0)(x - h) + (y - 0)(y - k) = 0  

x² + y² - hx - ky = 0

Answer 2

Let A( x, y) be any point on the Locus.

The given Straight line passes through B (h, k).

Since A, B are both the points on the same line, They are collinear and B is the foot of the perpendicular from O.

So, AB ⊥ AO .

$\sf \: Slope \: of \: AB = \frac{y - k}{x - h}$

$\sf \: Slope of AO = \frac{y - 0}{x - 0} = \frac{y}{x}$

Product of slopes of perpendicular lines is - 1

$\frac{y}{x} \times \frac{y - k}{x - h} = - 1 \\ \\ y(y - k) = - x(x - h) \\ \\ y ^{2} - yk \: = - {x}^{2} + xh \\ \\ {x}^{2} + {y}^{2} - xh - yk = 0$

This is required locus of foot of perpendicular drawn from origin to a variable straight line passing through (h, k).