Question

If $A(-2,1),B(2,3)$ and $C(-2,-4)$ are three points, find the angle between the lines $AB$ and $BC$.

Answer 1

Let the slope of line $AB$ and $BC$ be $m_1$ and $m_2$ respectively.

As we know that slope of line joining points $(x_1,y_1)$ and $(x_2,y_2)$ is given by-

$m=\frac{(y_2-y_1)}{(x_2-x_1)}$

Therefore,

$m_1=\frac{(3-1)}{(2-(-2))}=\frac{1}{2}$

$m_2=\frac{((-4)-3)}{((-2)-2)}=\frac{7}{4}$

Let $\theta$ be the angle between the lines $AB$ and $AC$.

$\therefore \tan \theta =\frac{m_2-m_1}{1+m_2{m}_1}$

$\Rightarrow \tan \theta =\frac{\left(\frac{7}{4}\right)-\left(\frac{1}{2}\right)}{1+\left(\frac{7}{4}\right)\left(\frac{1}{2}\right)}$

$\Rightarrow \tan \theta =\frac{14-4}{8+7}$

$\tan \theta =\frac{10}{15}=\frac{2}{3}$

$\Rightarrow \theta ={\tan }^{-1}\left(\frac{2}{3}\right)$

Hence angle between the lines $AB$ and $AC$ is ${\tan }^{-1}\left(\frac{2}{3}\right)$.