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 $BD$ is $x=b$.
Slope of $OB$ is $\frac{c}{b}$.
Slope of $AF$ is $-\frac{b}{c}$.
Now, the equation of altitude $AF$ is
$y-0=-\frac{b}{c}(x-a)$
Suppose, $BD$ and $OE$ intersect at $P$.
Coordinates of $P$ are $\left[b,b\left(\frac{(a-b)}{c}\right)\right]$
Let $m_1$ be the slope of $OP=\frac{a-b}{c}$
and $m_2$ be the slope of $AB=\frac{c}{b-a}$
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 $AB$.