Question

If a straight line $L$ is perpendicular to the line $4x-2y=1$ and forms a triangle of area $4$ square units with the coordinate axes, then an equation of the line $L$ is :

• A. $2x+4y+7=0$
• B. $2x-4y+8=0$
• C. $2x+4y+8=0$
• D. $4x-2y-8=0$

Answer 1

The correct option is C $2x+4y+8=0$

$L$ is perpendicular to $4x-2y=1\Rightarrow y=2x-1/2$

(Slope $P$)

(slope of $l$) $\times$(slope $P$) $=-1$

(slope of $L$) $\left(2\right)=-1$

Slope of $l=-1/2$

let line be of the form $y=mx+c$

$L\equiv y=\frac{-1}{2}x+c$

$x$ intercept of L:0$

$y=0\Rightarrow x=2c$

$y$intercept:

$x=0$ $\Rightarrow$ y=c$

area=1/2(xc)(c)=4....(given) $

$\Rightarrow c^2=4$

$\Rightarrow c=0+2$

$\therefore$ equation of line is y=\dfrac{-x}{2}+2=y=\dfrac{-x}{2}-2$

$\Rightarrow 2y+x=4$ or $2y+x+4=0$

$\Rightarrow 4y+2x-8=0$ or $4y+2x+8=0$