The length of the perpendicular from the point $P(a, b)$ to the line $\frac{x}{a}+\frac{y}{b}=1$ is
The length of the perpendicular from the point $P(a, b)$ to the line $\frac{x}{a}+\frac{y}{b}=1$ is
$\mid \frac{\sqrt{a^{2}+b^{2}}}{a b}$ units
$\left|\frac{a b}{\sqrt{a^{2}+b^{2}}}\right|$ units
$\left|\frac{b^{2}}{\sqrt{a^{2}+b^{2}}}\right|$ units
$\left|\frac{a^{2}}{\sqrt{a^{2}+b^{2}}}\right|$ units
Solution
We have line $b x+a y-a b=0$
Length of $\perp e r$ from $P(a, b)$ on the given line is
$\left|\frac{b a+a b-a b}{\sqrt{b^{2}+a^{2}}}\right|=\left|\frac{a b}{\sqrt{a^{2}+b^{2}}}\right|$