Question

Find the orthocentre of triangle whose vertices are $(2,0),(3,4)$ and $(0,3)$

Answer 1

Let, $A(2,0),B(3,4),C(0,3)$ be the vertices of $△ABC$ and $AD,BE,CF$ be the altitudes and $O(h,k)$ be the orthocentre of $△ABC$.

Now, $AD\perp BC\Rightarrow \left(slopeofAO\right)\times \left(slopeofBC\right)=-1\Rightarrow \left(\frac{k-0}{h-2}\right)\times \left(\frac{3-4}{0-3}\right)=-1\Rightarrow 6h+k=6⟶\left(1\right)$

Also, $BE\perp AC\Rightarrow \left(slopeofBO\right)\times \left(slopeofAC\right)=-1\Rightarrow \left(\frac{k-4}{h-3}\right)\times \left(\frac{3-0}{0-2}\right)=-1\Rightarrow -2h+3k=6⟶\left(2\right)$

Solving equations $\left(1\right)$ and $\left(2\right)$ we get,

$k=\frac{12}{5},h=\frac{3}{5}$

Hence, the orthocentre of $△ABC$ is $(\frac{12}{5},\frac{3}{5}).$