Q80 / 120 · VITEEE 2010 · Mathematics
If $\mathrm{A}(-2,1), \mathrm{B}(2,3)$ and $\mathrm{C}(-2,-4)$ are three points. Then, the angle between $\mathrm{BA}$ and $\mathrm{BC}$ is
- Correct answerA$\tan ^{-1}\left(\frac{2}{3}\right)$
- B$\tan ^{-1}\left(\frac{3}{2}\right)$
- C$\tan ^{-1}\left(\frac{1}{3}\right)$
- D$\tan ^{-1}\left(\frac{1}{2}\right)$
Solution
$A(-2,1), B(2,3)$ and $C(-2,-4)$ are three given points.
Slope of the line $B A$
$$
\begin{array}{l}
m_{1}=\frac{1-3}{-2-2}=\frac{1}{2} \\
\left(\text { Using slope formula, } m=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\right)
\end{array}
$$
Slope of the line $B C$
$$
m_{2}=\frac{-4-3}{-2-2}=\frac{7}{4}
$$
Now, angle between $A B$ and $B C$ is given by
$$
\begin{array}{l}
\tan \theta=\left|\frac{m_{1}-m_{2}}{1+m_{1} m_{2}}\right|=\left|\frac{\frac{1}{2}-\frac{7}{4}}{1+\frac{1}{2} \cdot \frac{7}{4}}\right| \\
\Rightarrow \tan \theta=\left|-\frac{10}{15}\right| \Rightarrow \tan \theta=\left|-\frac{2}{3}\right| \\
\Rightarrow \theta=\tan ^{-1}\left|-\frac{2}{3}\right| \Rightarrow \theta=\tan ^{-1}\left(\frac{2}{3}\right) \\
\quad[\because|-x|=x]
\end{array}
$$
Hence, the correct answer is option: 'A'.