Question

A straight line passes through a fixed point (h, k). Then the locus of the feet of the perpendiculars on it from the origin is

• A. +-hx-ky=0
• B. -hx+ky=0
• C. +hx+ky=0
• D. +-hx-ky=0

Answer 1

The correct option is A +-hx-ky=0

Let the equation of the straight line be
$xcos\alpha +ysin\alpha –p=0\dots ..\left(i\right)$
Since, the straight line (i) passes through the point (h, k), therefore
$hcos\alpha +ksin\alpha –p=0\dots ..\left(ii\right)$
On substracting Eq. (ii) from Eq. (i), we get
$(x-h)cos\alpha +(y–k)sin\alpha =0\dots ..\left(iii\right)$
Now, the equation of the straight line which is perpendicular to line (i) and passes through the origin, is
$xsin\alpha –ycos\alpha =0\dots ..\left(iv\right)$
The required locus will be obtained by eliminating α from the Eqs. (iii) and (iv). From Eq. (iv), $xsin\alpha =ycos\alpha$
$\Rightarrow \frac{x}{cos\alpha }=\frac{y}{sin\alpha }\Rightarrow \frac{x}{cos\alpha }=\frac{y}{sin\alpha }=\frac{\sqrt{x^2+y^2}}{\sqrt{co{s}^2\alpha +si{n}^2\alpha }}=\sqrt{x^2+y^2}\Rightarrow x(x-h)+y(y-k)=0\Rightarrow x^2-hx+y^2-ky=0\therefore x^2+y^2-hx-ky=0$
Which is the required locus.
Hence, (a) is the correct answer.