Question

If $f\left(\alpha \right)=x\cos \alpha +y\sin \alpha -p\left(\alpha \right)$, then the lines $f\left(\alpha \right)=0$ and $f\left(\beta \right)=0$ are perpendicular to each other, if

• A. $\alpha =\beta$
• B. $\alpha +\beta =\frac{\pi }{2}$
• C. $|\alpha -\beta |=\frac{\pi }{2}$
• D. $\alpha \pm \beta =\frac{\pi }{2}$

Answer 1

The correct option is C

$|\alpha -\beta |=\frac{\pi }{2}$

Explanation for the correct answer:

Step 1: Prerequisite

A general equation of lines is given by

$ax+by-c=0$ (i)

and its slope is given as

$m=\frac{-a}{b}$.

Also if two lines are perpendicular to each other, the relation between their slope is given as

$m_1{m}_2=-1$

where $m_1$ is the slope of the first line and $m_2$ is the slope of the second line.

Step 2: Making an assumption

Let us assume that the lines

$f\left(\alpha \right)=0$ and

$f\left(\beta \right)=0$

are perpendicular to each other and their slopes are given as $m_1$ and $m_2$ respectively.

Step 3: Evaluating the given data

From the given data,

$f\left(\alpha \right)=x\cos \alpha +y\sin \alpha -p\left(\alpha \right)$

It can be written as

$x\cos \alpha +y\sin \alpha -p\left(\alpha \right)=0\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\left(ii\right)$.

Similarly,

$f\left(\beta \right)=0$.

Therefore,

$x\cos \beta +y\sin \beta -p\left(\beta \right)=0\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\left(iii\right)$.

On comparing the given equations with equation (i), we get the slopes of the lines as

$m_1=\frac{-\cos \alpha }{\sin \alpha }$and

$m_2=\frac{-\cos \beta }{\sin \beta }$.

On substituting the values of slope in

$m_1{m}_2=-1$, we get

$$\begin{gathered} \frac{-\cos \alpha }{\sin \alpha }\times \frac{-\cos \beta }{\sin \beta }=-1 \\ \Rightarrow \cos \alpha \cos \beta =-\sin \alpha \sin \beta \\ \Rightarrow \cos \alpha \cos \beta +\sin \alpha \sin \beta =0 \\ \Rightarrow \cos (\alpha -\beta )=0 \\ [\because \cos \alpha \cos \beta -\sin \alpha \sin \beta =\cos (\alpha -\beta \left)\right] \\ \Rightarrow \cos (\alpha -\beta )=\cos \frac{\pi }{2} \\ [\because \cos \frac{\pi }{2}=0] \\ \Rightarrow \alpha -\beta =\frac{\pi }{2} \end{gathered}$$

Hence, Option (C) is the correct option.