Question

Find the coordinates of the orthocentre of the triangle whose vertices are (−1, 3), (2, −1) and (0, 0).

Answer 1

Let A (0, 0), B (−1, 3) and C (2, −1) be the vertices of the triangle ABC.
Let AD and BE be the altitudes.



$AD\perp BC$and $BE\perp AC$

$\therefore$ Slope of AD $\times$ Slope of BC = −1
Slope of BE $\times$ Slope of AC = −1

Here, slope of BC = $\frac{-1-3}{2+1}=-\frac{4}{3}$
and slope of AC = $\frac{-1-0}{2-0}=-\frac{1}{2}$

$$\begin{gathered} \therefore \mathrm{Slope}\mathrm{of}AD\times \left(-\frac{4}{3}\right)=-1\mathrm{and}\mathrm{slope}\mathrm{of}BE\times \left(-\frac{1}{2}\right)=-1 \\ \Rightarrow \mathrm{Slope}\mathrm{of}AD=\frac{3}{4}\mathrm{and}\mathrm{slope}\mathrm{of}BE=2 \end{gathered}$$

The equation of the altitude AD passing through A (0, 0) and having slope $\frac{3}{4}$ is

$$\begin{gathered} y-0=\frac{3}{4}\left(x-0\right) \\ \Rightarrow y=\frac{3}{4}x....\left(1\right) \end{gathered}$$

The equation of the altitude BE passing through B (−1, 3) and having slope 2 is

$$\begin{gathered} y-3=2\left(x+1\right) \\ \Rightarrow 2x-y+5=0....\left(2\right) \end{gathered}$$

Solving (1) and (2):
x = − 4, y = − 3

Hence, the coordinates of the orthocentre is (−4, −3).