Find the length of the altitudes of the triangle whose sides are the lines $3x - 4y = 5$, $4x + 3y = 5$ and $x + y = 1$.
The sides of the triangle are
$3x - 4y = 5$ $\;\;\; \cdots \; (1)$, $\;$ $4x + 3y = 5$ $\;\;\; \cdots \; (2)$, $\;$ $x + y = 1$ $\;\;\; \cdots \; (3)$
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 $\;$ $A \left(x_1, y_1\right) = \left(\dfrac{7}{5}, \dfrac{-1}{5}\right)$
Solving equations $(2)$ and $(3)$ simultaneously gives the point of intersection as $\;$ $B \left(x_2, y_2\right) = \left(2, -1\right)$
Solving equations $(3)$ and $(1)$ simultaneously gives the point of intersection as $\;$ $C \left(x_3, y_3\right) = \left(\dfrac{9}{7}, \dfrac{-2}{7}\right)$
Let $AD$ be the altitude drawn from point $A$ to line given by equation $(3)$.
Equation $(3)$ can be written as $\;$ $x + y - 1 = 0$
$\therefore \;$ Length of $AD = \dfrac{1 \times \dfrac{7}{5} + 1 \times \left(\dfrac{-1}{5}\right) - 1}{\sqrt{1^2 + 1^2}} = \dfrac{1}{5 \sqrt{2}}$
Let $BE$ be the altitude drawn from point $B$ to line given by equation $(1)$.
Equation $(1)$ can be written as $\;$ $3x - 4y - 5 = 0$
$\therefore \;$ Length of $BE = \dfrac{3 \times 2 - 4 \times \left(-1\right) - 5}{\sqrt{3^2 + \left(-4\right)^2}} = \dfrac{5}{5} = 1$
Let $CF$ be the altitude drawn from point $C$ to line given by equation $(2)$.
Equation $(2)$ can be written as $\;$ $4x + 3y - 5 = 0$
$\therefore \;$ Length of $CF = \dfrac{4 \times \dfrac{9}{7} + 3 \times \left(\dfrac{-2}{7}\right) - 5}{\sqrt{4^2 + 3^2}} = \dfrac{1}{5}$