Question

Using co-ordinate geometry, prove that the
three altitudes of any triangle are concurrent.

Answer 1

Choose one side $BC=2a$ along $x-$axis with its mid-point $O$ as origin so that the point $B$ and $C$ are $(-a,0)$ and $(a,0)$ respectively. Let the third vertex be the point $(h,k)$.
$AD,BE$ and $CF$ are three altitudes whose equations are
$ADx=h$ or $x+0.y-h=0$
$BEy-0=-\frac{(h-a)}{k}(x+a)$
Or $x(h-a)+ky+a(h-a)=0$
$CF$ Replace $a$ by $-a$ in $\left(2\right)$
Or $x(h+a)+ky-a(h+a)=0$
The three lines will be concurrent if $△=0$
Or $\left|\begin{array}{ccc}1& 0& -h\\ h-a& k& a(h-a)\\ h+a& k& -a(h+a)\end{array}\right|$
Apply $R_3-R_2$ and two rows become identical.
$\therefore △=0$ $\therefore$ concurrent.