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 altitutdes.

$AD\perp BC$ 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 = $\frac{-1-0}{2-0}=-\frac{1}{2}$

$\therefore$ Slope of AD $\times (-\frac{4}{3})=-1$ and slope of

$BE\times (-\frac{1}{2})=-1$

$\Rightarrow$ Slope of AD = $\frac{3}{4}$ and slope of BE = 2

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

$y-0=\frac{3}{4}(x-0)$

$\Rightarrow y=\frac{3}{4}x...\left(i\right)$

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

$y-3=2(x+1)$

$\Rightarrow 2x-y+5=0...\left(ii\right)$

Solving (i) and (ii):

$x=-4,y=-3$

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