Find the coordinates of the foot of the perpendicular from the point $\left(-2, 0\right)$ on the line $2x + 3y = 9$.
Given point $= A \left(x_1, y_1\right) = \left(-2, 0\right)$
Let the foot of the perpendicular from point $A$ be at the point $\;$ $B \left(x_2, y_2\right)$
Equation of given line is $\;$ $2x + 3y = 9$ $\;\;\; \cdots \; (1)$
Slope of the given line $\;$ $2x + 3y = 9$ is $\;$ $m = \dfrac{-2}{3}$
$\because \;$ $AB$ is perpendicular to the given line, slope of perpendicular $= m_1 = \dfrac{-1}{m} = \dfrac{3}{2}$
$\therefore \;$ Equation of perpendicular $AB$ passing through the point $A$ and having slope $= m_1$ is
$y - 0 = \dfrac{3}{2} \left(x + 2\right)$
i.e. $\;$ $2y = 3x + 6$
i.e. $\;$ $3x - 2y + 6 = 0$ $\;\;\; \cdots \; (2)$
Solving equations $(1)$ and $(2)$ simultaneously gives the point of intersection as $\left(0, 3\right)$
The point of intersection is the foot of the perpendicular from point $A$ on the line given by equation $(1)$.
$\therefore \;$ $B = \left(0, 3\right)$