Question

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

Answer 1

Given line $5x-y=1$

slope of line $=\frac{-5}{-1}=5$

product of slopes of two perpendicular

lines is - 1.

$m_1\times 5=-1$

slope of required line $m_1=\frac{-1}{5}$

Let The equation of required line be

$\Rightarrow y=mx+c$ $[m\to slope]$

$m=\frac{-1}{5}\Rightarrow y=\frac{-x}{5}+c$

Area of triangle formed by a line

with intercepts $a,b=\frac{1}{2}ab$

Here x - intercept = 5c

y - intercept = c

Area of $△le=\frac{1}{2}\left(5c\right)\left(c\right)=\frac{5c^2}{2}$

Given Area = 5 $\therefore \frac{5c^2}{2}=5\Rightarrow c^2=2$

$c=\pm \sqrt{2}$

$\therefore$ The equation of line $y=\frac{-x}{5}\pm \sqrt{2}$