Question

If the equation $y=mx+c$ and $x\cos \left(\alpha \right)+y\sin \left(\alpha \right)=p$ represents the same straight line, then

• A. $p=c\sqrt{(1+m^2)}$
• B. $c=p\sqrt{(1+m^2)}$
• C. $pc=\sqrt{(1+m^2)}$
• D. $p^2+c^2+m^2=1$

Answer 1

The correct option is B

$c=p\sqrt{(1+m^2)}$

Explanation for correct option

The given equation of the line is

$y=mx+c....\left(i\right)$

$x\cos \left(\alpha \right)+y\sin \left(\alpha \right)=p.....\left(ii\right)$

Equation $\left(ii\right)$ can be written as

$$\begin{gathered} x\cos \left(\alpha \right)+y\sin \left(\alpha \right)=p \\ \Rightarrow y\sin \left(\alpha \right)=p-x\cos \left(\alpha \right) \\ \Rightarrow y=-\frac{x\cos \left(\alpha \right)}{\sin \left(\alpha \right)}+\frac{p}{\sin \left(\alpha \right)} \\ \Rightarrow y=-x\cot \left(\alpha \right)+\frac{p}{\sin \left(\alpha \right)} \end{gathered}$$

Now comparing with equation $\left(i\right)$ we have

$m=-\cot \left(\alpha \right),c=\frac{p}{\sin \left(\alpha \right)}$

From $c=\frac{p}{\sin \left(\alpha \right)}$ we have $\sin \left(\alpha \right)=\frac{p}{c}........\left(iii\right)$

From $m=-\cot \left(\alpha \right)$ we have

$$\begin{gathered} \Rightarrow m=-\cot \left(\alpha \right) \\ \Rightarrow m=-\frac{\cos \left(\alpha \right)}{\sin \left(\alpha \right)} \\ \Rightarrow m\sin \left(\alpha \right)=-\cos \left(\alpha \right) \\ \Rightarrow \frac{mp}{c}=-\cos \left(\alpha \right)\left[\because \sin \left(\alpha \right)=\frac{p}{c}\right]......\left(iv\right) \end{gathered}$$

Now,

$$\begin{gathered} {\cos }^2\left(\alpha \right)+{\sin }^2\left(\alpha \right)=1 \\ \Rightarrow {\left(\frac{-mp}{c}\right)}^2+{\left(\frac{p}{c}\right)}^2=1\left[\mathrm{From}\left(\mathrm{iii}\right)\mathrm{and}\left(\mathrm{iv}\right)\right] \\ \Rightarrow \frac{m^2{p}^2+p^2}{c^2}=1 \\ \Rightarrow p^2\left(m^2+1\right)=c^2 \\ \Rightarrow p\sqrt{m^2+1}=c \end{gathered}$$

Hence, option (B) is correct i.e. $c=p\sqrt{(1+m^2)}$