Question

Find the equations of the altitudes of a ∆ ABC whose vertices are A (1, 4), B (−3, 2) and C (−5, −3).

Answer 1

The vertices of ∆ABC are A (1, 4), B (−3, 2) and C (−5, −3).


Slope of AB = $\frac{2-4}{-3-1}=\frac{1}{2}$

Slope of BC = $\frac{-3-2}{-5+3}=\frac{5}{2}$

Slope of CA = $\frac{4+3}{1+5}=\frac{7}{6}$

Thus, we have:

Slope of CF = $-2$

Slope of AD = $-\frac{2}{5}$

Slope of BE = $-\frac{6}{7}$

Hence,

$$\begin{gathered} \mathrm{Equation}\mathrm{of}\mathrm{CF}\mathrm{is}: \\ y+3=-2\left(x+5\right) \\ \Rightarrow 2x+y+13=0 \\ \mathrm{Equation}\mathrm{of}\mathrm{AD}\mathrm{is}: \\ y-4=-\frac{2}{5}\left(x-1\right) \\ \Rightarrow 2x+5y-22=0 \\ \mathrm{Equation}\mathrm{of}\mathrm{BE}\mathrm{is}: \\ y-2=-\frac{6}{7}\left(x+3\right) \\ \Rightarrow 6x+7y+4=0 \end{gathered}$$