If $\left(x_1, y_1\right)$ be the coordinates of the foot of the perpendicular from the origin to the line $\dfrac{x}{a} + \dfrac{y}{b} = 1$, show that $\left(x_1^2 + y_1^2\right) \left(x_1 + y_1\right) = \left(a + b\right) x_1 y_1$
Let $A = \left(0, 0\right)$; $\;$ $B = \left(x_1, y_1\right)$ $\;$ be the foot of the perpendicular
Slope of perpendicular $AB = m_1 = \dfrac{y_1}{x_1}$
Equation of given line is $\;$ $\dfrac{x}{a} + \dfrac{y}{b} = 1$
i.e. $\;$ $bx + ya = ab$
$\therefore \;$ Slope of given line $= m_2 = \dfrac{-b}{a}$
$\because \;$ $AB$ is perpendicular to the given line, $\;$ $m_1 = \dfrac{-1}{m_2}$
i.e. $\;$ $\dfrac{y_1}{x_1} = \dfrac{a}{b}$ $\implies$ $y_1 = \dfrac{a x_1}{b}$ $\;\;\; \cdots \; (1)$
$\because \;$ $B$ is the foot of the perpendicular, point $B$ lies on the given line.
$\therefore \;$ We have, $\;$ $\dfrac{x_1}{a} + \dfrac{y_1}{b} = 1$
i.e. $\;$ $bx_1 + ay_1 = ab$
i.e. $\;$ $bx_1 + a \times \dfrac{ax_1}{b} = ab$ $\;\;\;$ [in view of equation $(1)$]
i.e. $\;$ $b^2 x_1 + a^2 x_1 = ab^2$
i.e. $\;$ $\left(a^2 + b^2\right) x_1 = ab^2$ $\implies$ $a^2 + b^2 = \dfrac{ab^2}{x_1}$ $\;\;\; \cdots \; (2)$
Now,
$\begin{aligned}
\left(x_1^2 + y_1^2\right) \left(x_1 + y_1\right) & = \left(x_1^2 + \dfrac{a^2 x_1^2}{b^2}\right) \left(x_1 + \dfrac{ax_1}{b}\right) \\\\
& = \dfrac{\left(b^2 x_1^2 + a^2 x_1^2\right) \left(b x_1 + a x_1\right)}{b^3} \\\\
& = \dfrac{x_1^2 \left(a^2 + b^2\right) x_1 \left(a + b\right)}{b^3} \\\\
& = \dfrac{x_1^3 \left(a^2 + b^2\right) \left(a + b\right)}{b^3} \\\\
& = x_1^3 \times \dfrac{ab^2}{x_1} \times \dfrac{\left(a + b\right)}{b^3} \;\;\; \left[\text{by equation (2)}\right] \\\\
& = \dfrac{a \left(a + b\right)x_1^2}{b} \;\;\; \cdots \; (3)
\end{aligned}$
$\begin{aligned}
\left(a + b\right) x_1 y_1 & = \left(a + b\right) \times x_1 \times \dfrac{ax_1}{b} \\\\
& = \dfrac{a \left(a + b\right) x_1^2}{b} \;\;\; \cdots \; (4)
\end{aligned}$
$\therefore \;$ We have from equations $(3)$ and $(4)$
$\left(x_1^2 + y_1^2\right) \left(x_1 + y_1\right) = \left(a + b\right) x_1 y_1$
Hence proved.