Question

Find the coordinates of the orthocentre of the triangles whose angular points are
$(1,0),(2,-4)$, and $(-5,-2)$.

Answer 1

Slope of $AB$ $=m_{AB}=\frac{0-(-2)}{1-(-5)}=\frac{1}{3}$

Draw $CE$ perpendicular $AB$

$m_{AB}{m}_{CE}=-1\frac{1}{3}{m}_{CE}=-1\Rightarrow m_{CE}=-3$

Equation of $CE$ is

$y+4=-3(x-2)y+4=-3x+6y+3x=\mathrm{2.......}\left(i\right)$

Now slope of $BC$ $=m_{BC}=\frac{-4-0}{2-1}=-4$

Draw $AD$ perpendicular $BC$

$m_{BC}{m}_{AD}=-1-4m_{AD}=-1\Rightarrow m_{AD}=\frac{1}{4}$

Equation of $AD$ is

$y+2=\frac{1}{4}(x+5)4y+8=x+5x-4y=\mathrm{3......}\left(ii\right)$

Orthocentre is the point of intersection of perpendicular from opposite vertices .So it is the point of intersection of $AD$ and $CE$

Solving $\left(i\right)$ and $\left(ii\right)$

$\Rightarrow x=\frac{11}{13},y=-\frac{7}{13}$

So the orthocentre of triangle is $(\frac{11}{13},-\frac{7}{13})$