The equations of the sides of a triangle are $\;$ $x - y - 1 = 0$, $\;$ $2x - 5 = 0$ $\;$ and $\;$ $3x + y -2 = 0$. Find the coordinates of the orthocenter.
The sides of the triangle are
$x - y -1 = 0$ $\;\;\; \cdots \; (1)$, $\;$ $2x - 5 = 0$ $\;\;\; \cdots \; (2)$, $\;$ $3x + y -2 = 0$ $\;\;\; \cdots \; (3)$
Let equations $(1)$ and $(2)$ intersect at point $A$.
Let equations $(2)$ and $(3)$ intersect at point $B$.
Let equations $(3)$ and $(1)$ intersect at point $C$.
Solving equations $(1)$ and $(2)$ simultaneously gives the point of intersection as $\;$ $A \left(x_1, y_1\right) = \left(\dfrac{5}{2}, \dfrac{3}{2}\right)$
Solving equations $(2)$ and $(3)$ simultaneously gives the point of intersection as $\;$ $B \left(x_2, y_2\right) = \left(\dfrac{5}{2}, \dfrac{-11}{2}\right)$
Solving equations $(3)$ and $(1)$ simultaneously gives the point of intersection as $\;$ $C \left(x_3, y_3\right) = \left(\dfrac{3}{4}, \dfrac{-1}{4}\right)$
Let $AD$ be the altitude drawn from point $A$ to line given by equation $(3)$.
Slope of equation $(3)$ is $m_1 = -3$
$\therefore \;$ Slope of altitude $AD = \dfrac{-1}{m_1} = \dfrac{1}{3}$
$AD$ passes through the point $A$ and has slope $\dfrac{-1}{m_1}$.
$\therefore \;$ Equation of $AD$ is
$y - \dfrac{3}{2} = \dfrac{1}{3} \left(x - \dfrac{5}{2}\right)$
i.e. $\;$ $6y - 9 = 2x - 5$
i.e. $\;$ $x - 3y + 2 = 0$ $\;\;\; \cdots \; (4)$
Let $BE$ be the altitude drawn from point $B$ to line given by equation $(1)$.
Slope of equation $(1)$ is $m_2 = 1$
$\therefore \;$ Slope of altitude $BE = \dfrac{-1}{m_2} = -1$
$BE$ passes through the point $B$ and has slope $\dfrac{-1}{m_2}$.
$\therefore \;$ Equation of $BE$ is
$y + \dfrac{11}{2} = -1 \left(x - \dfrac{5}{2}\right)$
i.e. $\;$ $2y + 11 = -2x + 5$
i.e. $\;$ $x + y + 3 = 0$ $\;\;\; \cdots \; (5)$
Solving equations $(4)$ and $(5)$ simultaneously gives the point of intersection as $\left(\dfrac{-11}{4}, \dfrac{-1}{4}\right)$
$\therefore \;$ The orthocenter is $\left(\dfrac{-11}{4}, \dfrac{-1}{4}\right)$.