Find the orthocenter of the triangle whose vertices are (5,–2), (–1,2), (1,4).

Given A = (5,–2), B = (–1,2), C = (1,4)

$\text{ Slope of }\overline{\mathrm{BC}}=\frac{{\mathrm{y}}_2-{\mathrm{y}}_1}{{\mathrm{x}}_2-{\mathrm{x}}_1}=\frac{4-2}{+1+1}=\frac{2}{2}=1$
$\begin{array}{l} Since \overline{AD}\perp \overline{\mathrm{BC} } then slope of\\ then slope of \overline{AD}=-1 \\ Equation of \overline{AD} is \mathrm{y}-{\mathrm{y}}_1=\mathrm{m}\left(\mathrm{x}-{\mathrm{x}}_1\right)\\ \mathrm{where} \mathrm{A}=(5,-2);\mathrm{m}=-1\Rightarrow \mathrm{y}+2=-1(\mathrm{x}-5)\end{array}$
$\mathrm{y}+2+\mathrm{x}-5=0;\mathrm{x}+\mathrm{y}-3=0\dots \left(1\right)$
$\text{ slope of }\overline{\mathrm{AC}}=\frac{{\mathrm{y}}_2-{\mathrm{y}}_1}{{\mathrm{x}}_2-{\mathrm{x}}_1}=\frac{4+2}{1-5}=\frac{6}{-4}=-3/2$
$\overline{\mathrm{AC}}\perp \overline{\mathrm{BE}}\Rightarrow \text{ slope of }\overline{\mathrm{BE}}=\frac{-1}{\frac{-3}{2}}=\frac{2}{3}$
$\text{ Equation of }\overline{\mathrm{BE}}\Rightarrow \mathrm{y}-{\mathrm{y}}_1=\mathrm{m}\left(\mathrm{x}-{\mathrm{x}}_1\right)$
$\begin{array}{l}\mathrm{B}=(-1,2)\Rightarrow \mathrm{y}-2=\frac{2}{3}(\mathrm{x}+1)\\ \Rightarrow 3\mathrm{y}-6=2\mathrm{x}+2\\ \Rightarrow 2\mathrm{x}-3\mathrm{y}+8=0\dots \left(2\right)\end{array}$
solving (1) & (2)
$$\begin{gathered} \mathrm{x} \mathrm{y} 1 \\ 1 -3 1 1 \\ -3 8 2 -3 \end{gathered}$$
$\begin{array}{l}\frac{\mathrm{x}}{8-9}=\frac{\mathrm{y}}{-6-8}=\frac{1}{-3-2}\\ \frac{\mathrm{x}}{-1}=\frac{\mathrm{y}}{-14}=\frac{1}{-5}\Rightarrow \frac{\mathrm{x}}{-1}=\frac{1}{-5}\Rightarrow \mathrm{x}=\frac{1}{5}\\ \frac{\mathrm{y}}{-14}=\frac{1}{-5}\Rightarrow \mathrm{y}=\frac{14}{5}\end{array}$
$\text{ Orthocenter }=\left(\frac{1}{5},\frac{14}{5}\right)$