Question

A straight line L is perpendicular to the line $5x-y=1$. The area of the triangle formed by the line coordinate axes is $5$. Find the equation of the line.

Answer 1

Let the linebe $\frac{x}{a}+\frac{y}{b}=1$

Slope $=-\frac{b}{a}$

it is perpendicular to $5x-1=y$

$\Rightarrow$ $m_1\times m_2=-1$

$\Rightarrow$ $5\times -\frac{b}{a}=-1$

$\Rightarrow$ $\frac{b}{a}=\frac{1}{5}\Rightarrow$ $a=56$

$\Rightarrow$ $\frac{x}{a}+\frac{y}{b}=1\Rightarrow$ $\frac{x}{5b}+\frac{y}{b}=1$

Given area with coordinate axes $=5$

$\Rightarrow$ $\frac{1}{2}\times ab=5$

$ab=10$

$5b\times b=10$

and $b^2=2\Rightarrow$ $b=\sqrt{2}$

$a=5b=5\sqrt{2}$

Equation of the line is

$\frac{x}{5\sqrt{2}}+\frac{y}{\sqrt{2}}=1\Rightarrow x+5y=5\sqrt{2}$