Question

Equation $\frac{x-x_1}{\cos \theta }=\frac{y-y_1}{\sin \theta }=r$ , represents-

• A. Equation of straight line, if $\theta$ is constant and r is variable .
• B. Equation of circle, if r is constant and $\theta$ is variable.
• C. A straight line passing through a fixed point and and having known slope.
• D. A circle with a known centre and given radius.

Answer 1

The correct option is A Equation of straight line, if $\theta$ is constant and r is variable .

Given

$\frac{x-x_1}{\cos \theta }=\frac{y-y_1}{\sin \theta }$

Now ,

$\frac{x-x_1}{\cos \theta }=\frac{y-y_1}{\sin \theta }$

$\frac{\sin \theta }{\cos \theta }=\frac{y-y_1}{x-x_1}$

$\tan \theta =\frac{y-y_1}{x-x_1}$

We know that slop of a line

$m=\tan \theta$

$\therefore m=\frac{y-y_1}{x-x_1}$

Hence this is the slop of line

Now given that

$\frac{y-y_1}{x-x_1}=r$ which is variable because x,y are variable

Hence option A is correct