Question

Find the coordinates of the orthocentre of the triangles whose angular points are
$(0,0),(2,-1)$, and $(-1,3)$.

Answer 1

Slope of $AC$ $=m=\frac{3-0}{-1-0}=-3$

Draw $BD$ perpendicular to $AC$ and its slope be $m^{\prime }$

$m{m}^{\prime }=-1-3m^{\prime }=-1\Rightarrow m^{\prime }=\frac{1}{3}$

Equation of $BD$ is

$y+1=\frac{1}{3}(x-2)3y+3=x-2x-3y=\mathrm{5.........}\left(i\right)$

Now slope of $AB$ $=m_{AB}=\frac{-1-0}{2-0}=-\frac{1}{2}$

Draw $CE$ perpemdicular $AB$

$=m_{AB}=\frac{-1-0}{2-0}=-\frac{1}{2}{m}_{AB}{m}_{CE}=-1-\frac{1}{2}{m}_{CE}=-1\Rightarrow m_{CE}=2$

Equation of $CE$ is

$y-3=2(x+1)2x-y+5=\mathrm{0......}\left(ii\right)$

Orthocentre is the point of intersection of perpendiculars drawn from opposite vertices . So it is the point of intersection of $BD$ and $CE$

Solving $\left(i\right)$ and $\left(ii\right)$ by substituting $x$ from $\left(i\right)$ in $\left(ii\right)$

$y-3=2(x+1)2(3y+5)-y+5=06y+10-y+5=05y+15=0\Rightarrow y=-3x=3(-3)+5\Rightarrow x=-4$

So the orthocentre of the triangle is $(-4,-3)$