Find the orthocentre of the triangle with the vertices (–2, –1), (6, –1) and (2, 5).

$$\begin{gathered} Given vertices of triangle are A=(-2,-1),B=(6,-1),C=(2,5) \\ Now \text{ Slope of }\overline{\mathrm{BC}}=\frac{-1-5}{6-2}=-\frac{3}{2} \\ \Rightarrow \text{ slope of }\mathrm{AD}\perp \mathrm{BC}=\frac{2}{3} \end{gathered}$$

$\text{ Equation to }\overline{\mathrm{AD}}\text{ is }\mathrm{y}+1=\frac{2}{3}(\mathrm{x}+2)$
$\begin{array}{l}\Rightarrow 2\mathrm{x}-3\mathrm{y}+1=0\\ \Rightarrow \text{ slope of }\overline{\mathrm{AC}}=\frac{-1-5}{-2-2}=\frac{3}{2}\\ \Rightarrow \text{ slope of }\overline{\mathrm{BE}}\perp \mathrm{AC}=-\frac{2}{3}\end{array}$
$\therefore \text{ Equation to }\overline{\mathrm{BE}}\text{ is }\mathrm{y}+1=-\frac{2}{3}(\mathrm{x}-6)$
$\Rightarrow 2\mathrm{x}+3\mathrm{y}-9=0 .....\left(2\right)$
Solving (1) and (2), the orthocenter of $∆$ ABC is $\mathrm{H}=\left(2,\frac{5}{3}\right)$