Q98 / 160 · AP EAMCET 2013 · Mathematics
The perpendicular distance from the point $(1, \pi)$ to the line joining $\left(1,0^{\circ}\right)$ and $\left(1, \frac{\pi}{2}\right)$, (in polar coordinates) is
- A$2$
- B$\sqrt{3}$
- C$1$
- Correct answerD$\sqrt{2}$
Solution
Given points $(1, \pi),\left(1,0^{\circ}\right)$ and $\left(1, \frac{\pi}{2}\right)$ are in polar form.
Now, change in cartesian form,
$$
\begin{aligned}
& (1, \pi) \rightarrow(1 \cdot \cos \pi, 1 \cdot \sin \pi) \rightarrow(-1,0) \\
& \left(1,0^{\circ}\right) \rightarrow\left(1 \cdot \cos 0^{\circ}, 1 \cdot \sin 0^{\circ}\right) \rightarrow(1,0) \\
& \text { and }\left(1, \frac{\pi}{2}\right) \rightarrow\left(1 \cdot \cos \frac{\pi}{2}, 1 \cdot \sin \frac{\pi}{2}\right) \rightarrow(0,1) \\
&
\end{aligned}
$$
Now, equation of the line passing through $(1,0)$ and $(0,1)$ is
$$
\begin{aligned}
& (y-0)=\frac{1-0}{0-1}(x-1) \\
& \Rightarrow \quad y=-x+1 \\
& \Rightarrow \quad x+y-1=0 \\
&
\end{aligned}
$$
So, the perpendicular distance from the point $(-1,0)$ to the line (i) is
$$
=\frac{|-1+0-1|}{\sqrt{1+1}}=\frac{2}{\sqrt{2}}=\sqrt{2}
$$
Hence, the correct answer is option: 'D'.