Question

Using coordinate geometry, prove that the three altitudes of any triangle are concurrent.

Answer 1

Let the vertices of a triangle be, $O(0,0) A(a, 0)$ and $B(b, c)$ equation of altitude $B D$ is $x=b$.
Slope of $O B$ is $\frac{c}{b}$.
Slope of $A F$ is $-\frac{b}{c}$.
Now, the equation of altitude $A F$ is
$y-0=-\frac{b}{c}(x-a)$
Suppose, $B D$ and $O E$ intersect at $P$.
Coordinates of $P$ are $\left[b, b\left(\frac{(a-b)}{c}\right)\right]$
Let $m_{1}$ be the slope of $O P=\frac{a-b}{c}$
and $ m_{2}$ be the slope of $A B=\frac{c}{b-a}$
Now, $ m_{1} m_{2}=\left(\frac{a-b}{c}\right)\left(\frac{c}{b-a}\right)=-1$
We get, that the line through $O$ and $P$ is perpendicular to $A B$.