If the equations of the sides of a triangle are 7x + y –10 = 0, x – 2y + 5 = 0 and x + y + 2 = 0. Find the orthocentre of the triangle.

Let the given triangle ABC with the sides $\overline{\mathrm{AB}},\overline{\mathrm{BC}}\mathrm{\&}\overline{\mathrm{CA}} \mathrm{are}$
$\begin{array}{l}\mathrm{x}-2\mathrm{y}+5=0 ......\left(1\right)\\ 7\mathrm{x}+\mathrm{y}-10=0 ......\left(2\right)\\ \text{ and }\mathrm{x}+\mathrm{y}+2=0 ......\left(3\right)\end{array}$

Given sides
Solve (1) and (3) A = (–3, 1)
Solve (1) and (2) B = (1, 3)
Solve (2) and (3) C = (2, – 4)
$$\begin{gathered} \text{ solpe of }\overline{\mathrm{BC}}=-7 \\ \mathrm{Let} \overline{\mathrm{AD}}\perp \overline{\mathrm{BC}} \\ slope of \overline{\mathrm{AD}}=\frac{1}{7} \end{gathered}$$
Equation of AD is
$\begin{array}{l}\mathrm{y}-{\mathrm{y}}_1=\frac{-1}{\mathrm{m}}\left(\mathrm{x}-{\mathrm{x}}_1\right)\Rightarrow \mathrm{y}-1=\frac{1}{7}(\mathrm{x}+3)\\ \Rightarrow \mathrm{x}-7\mathrm{y}+10=0\dots \dots \text{ (4) }\end{array}$
$\text{ slope of }\overline{\mathrm{AC}}=-1$

$\text{ and }\overline{\mathrm{BE}}\perp \overrightarrow{\mathrm{AC}}$
eq’n of BE is y – 3 = 1 (x – 1)
$\mathrm{x}-\mathrm{y}+2=0 .....\left(5\right)$
$\mathrm{Solve} \left(4\right) \mathrm{and} \left(5\right) (\mathrm{x},\mathrm{y})=\left(\frac{-2}{3},\frac{4}{3}\right)$
$\therefore \text{ Orthocentre of triangle }\mathrm{ABC}\text{ is }\left(\frac{-2}{3},\frac{4}{3}\right)$