The orthocenter of the triangle formed by (1, -3), (6, 1), (4, -1) is

Let  A (1, -3)   B (6, 1)  C (4, -1)

$\begin{array}{l}Slopeof BC=\frac{-1-1}{4-6}=\frac{-2}{-2}=1\\ AD\perp \overline{BC}andA\left(1,-3\right)=\left(x_1,y_1\right)\\ \Rightarrow Equationof\overline{AD}is\\ \Rightarrow y+3=\frac{-1}{1}\left(x-1\right)\\ \Rightarrow y+3=-x+1\Rightarrow x+y+2=0--\left(1\right)\\ Slopeof\overline{AC}=\frac{-1+3}{4-1}=\frac{2}{3}\\ \overline{BE}\perp \overline{AC}andB\left(6,1\right)=\left(x_1,y_1\right)\end{array}$

$\begin{array}{l}Equationof\overline{BE}isy-1=\frac{-1}{\left(\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.\right)}\left(x-6\right)\\ \Rightarrow y-1=\frac{-3}{2}\left(x-6\right)\\ \Rightarrow 2y-2=-3x+18\\ \Rightarrow 3x+2y-20=0--\left(2\right)\end{array}$

Solving  (1) & (2)

$\begin{array}{l}xy1\\ \begin{array}{cccc}1& 2& 1& 1\\ 2& -20& 3& 2\end{array}\\ \Rightarrow \frac{x}{-20-4}=\frac{y}{6+20}=\frac{1}{2-3}\\ \Rightarrow \frac{x}{-24}=\frac{y}{26}=\frac{1}{-1}\\ \therefore x=24,y=-26\\ \therefore orthocentre=\left(24,-26\right)\end{array}$