Straight Line Question 138

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

Options:

A) $ x^{2}+y^{2}-hx-ky=0 $

B) $ x^{2}+y^{2}+hx+ky=0 $

C) $ 3x^{2}+3y^{2}+hx-ky=0 $

D)None of these

Show Answer

Answer:

Correct Answer: A

Solution:

  • $ y-k=m(x-h) $ and $ y-0=-\frac{1}{m}(x-0) $ . Eliminate m and replace (h,k) by $ (x,y) $ , we get $ x^{2}+y^{2}-hx-ky=0 $ , which is the required locus of the point.