Question

Find the length of perpendicular drawn from (b,a) to the line x/a + y/b =1

Answer 1

Step-by-step explanation:

Given equation of line is

$\frac{ \times }{a} \: + \: \frac{y}{b} = 1$

$We \: written \: that \: this \: equation bx+ay=ab …… (1)$

$We \: know \: that$

$Length \: of \: perpendicular \: any \: line \\ \: ax+by+c=0$

$d \: = \frac{ax + by + c}{ \sqrt{a {}^{2} } + \: b {}^{2} }$

$From \: equation \: (1),$

$bx + ay \: - ab = 0$

$Length \: of \: perpendicular \: is$

$d \: = \frac{bx \: + ay \: - ab}{ \sqrt{b {}^{2} + a {}^{2} } }$

$Length \: of \: perpendicular \: given \: point \: \\ (a,b)$

$d \: = \frac{b \times a + a \times b - ab}{ \sqrt{b {}^{2} + ab {}^{2} } }$

$d \: = \frac{ab}{ \sqrt{a {}^{2} + b {}^{2} } }$

Hence it is required solution.