Question

If $A(-2,1),B(2,3)$ and $C(-2,-4)$ are three points, then the acute angle between the lines $BA$ and $BC$ is

• A. ${\tan }^{-1}\left(\frac{2}{3}\right)$
• B. ${\tan }^{-1}\left(\frac{4}{3}\right)$
• C. ${\tan }^{-1}\left(\frac{1}{3}\right)$
• D. ${\tan }^{-1}\left(\frac{3}{2}\right)$

Answer 1

The correct option is A ${\tan }^{-1}\left(\frac{2}{3}\right)$

Let $m_1$ and $m_2$ be the slopes of $BA$ and $BC$, respectively. Then,
$m_1=\frac{3-1}{2-(-2)}=\frac{2}{4}=\frac{1}{2}$
and $m_2=\frac{-4-3}{-2-2}=\frac{7}{4}$
If $\theta$ is the acute angle between them, then
$\tan \theta =\left|\frac{m_2-m_1}{1+m_1{m}_2}\right|\Rightarrow \tan \theta =\left|\frac{\frac{7}{4}-\frac{1}{2}}{1+\frac{7}{4}\times \frac{1}{2}}\right|\Rightarrow \tan \theta =\left|\frac{\frac{10}{8}}{\frac{15}{8}}\right|=\frac{2}{3}$
$\Rightarrow \theta ={\tan }^{-1}\left(\frac{2}{3}\right)$