The equation of a given straight line is $\frac{x-x_1}{\cos \theta}=\frac{y-y_1}{\sin \theta}=\gamma$. If the equation of the line perpendicular to the given line and passing through $(\alpha, \beta)$ is $\frac{x}{a}+\frac{y}{b}=1$, then $\frac{b}{a}$ is equal to

• A

  $\tan \theta$
• B

  $-\tan \theta$
• Correct answer
  C

  $\cot \theta$
• D

  $-\cot \theta$

Solution

Given equation of line is $\frac{x-x_1}{\cos \theta}=\frac{y-y_1}{\sin \theta}=\gamma$ Slope of Eq. (i) is $m=\tan \theta$
So, slope of a line perpendicular to Eq. (i) is $(-\cot \theta)$. According to the question, equation of required line is $\frac{x}{a}+\frac{y}{b}=1$ Slope of line (ii) is $\left(\frac{-b}{a}\right)$ So, $\frac{-b}{a}=-\cot \theta \Rightarrow \frac{b}{a}=\cot \theta$

Hence, the correct answer is option: 'C'.