Question

Find the equations to the altitudes of the triangle whose angular points are A (2, −2), B (1, 1) and C (−1, 0).

Answer 1

Let $m_{AD},m_{BE}\mathrm{and}{m}_{CF}$ be the slopes of the altitudes AD, BE and CF, respectively.


$$\begin{gathered} \therefore \mathrm{Slope}\mathrm{of}AD\times \mathrm{Slope}\mathrm{of}BC=-1 \\ \Rightarrow m_{AD}\times \left(\frac{0-1}{-1-1}\right)=-1 \\ \Rightarrow m_{AD}\times \frac{1}{2}=-1 \\ \Rightarrow m_{AD}=-2 \end{gathered}$$


$$\begin{gathered} \mathrm{Slope}\mathrm{of}BE\times S\mathrm{lope}\mathrm{of}AC=-1 \\ \Rightarrow m_{BE}\times \left(\frac{0+2}{-1-2}\right)=-1 \\ \Rightarrow m_{BE}\times \left(\frac{-2}{3}\right)=-1 \\ \Rightarrow m_{BE}=\frac{3}{2} \end{gathered}$$


$$\begin{gathered} \mathrm{Slope}\mathrm{of}CF\times S\mathrm{lope}\mathrm{of}AB=-1 \\ \Rightarrow m_{CF}\times \left(\frac{1+2}{1-2}\right)=-1 \\ \Rightarrow m_{CF}\times \left(-3\right)=-1 \\ \Rightarrow m_{CF}=\frac{1}{3} \end{gathered}$$

Now, the equation of AD which passes through A (2, −2) and has slope −2 is

$$\begin{gathered} y+2=-2\left(x-2\right) \\ \Rightarrow 2x+y-2=0 \end{gathered}$$

The equation of BE, which passes through B (1, 1) and has slope $\frac{3}{2}$ is
$$\begin{gathered} y-1=\frac{3}{2}\left(x-1\right) \\ \Rightarrow 3x-2y-1=0 \end{gathered}$$

The equation of CF, which passes through C (−1, 0) and has slope $\frac{1}{3}$ is
$$\begin{gathered} y-0=\frac{1}{3}\left(x+1\right) \\ \Rightarrow x-3y+1=0 \end{gathered}$$