The given vertices are A(x1,y1) = (– 2, 1),
B(x2,y2) = (2, 3) and C(x3,y3) = (– 2, – 4).
Slope of AB, m1 = \(\frac{y_2-y_1}{x_2-x_1}\)
=\(\frac{3-1}{2-(-2)}\)
=\(\frac{1}{2}\)
And slope of BC, m2 = \(\frac{y_3-y_2}{x_3-x_2}\)
=\(\frac{-4-3}{-2-2}\)
=\(\frac{7}{4}\)
∴ Angle B, which is the angle between AB and BC is given by
\(\theta\) = tan-1 \(\frac{m_2-m_1}{1+m_1m_2}\)
= tan-1 \(\frac{\frac{7}{4}-\frac{1}{2}}{1+\frac{7}{4}.\frac{1}{2}}\)
= tan-1 \((\frac{5}{4}\times\frac{8}{15})\)
= tan-1 \((\frac{2}{3})\).
