Question

Let line $L$ passes through the point of intersection of $2x+y-1=0$ and $x+2y-2=0$. If $L$ makes a triangle with the coordinate axes of area $\frac{5}{8}\text{sq. units}$, then the equation of $L$ can be

• A. $4x-5y+5=0$
• B. $4x+5y-5=0$
• C. $4x-5y-5=0$
• D. $4x+5y+5=0$

Answer 1

Answer: (A), (B)

$4x+5y-5=0$

The equation of $L$ is
$2x+y-1+\lambda (x+2y-2)=0$
$\Rightarrow (2+\lambda )x+(1+2\lambda )y-1-2\lambda =0$
Dividing by $1+2\lambda$, we get
$\frac{x}{\frac{1+2\lambda }{2+\lambda }}+\frac{y}{1}-1=0$

$x-$ intercept is $\frac{1+2\lambda }{2+\lambda }$
$y-$ intercept is $1$

So, the area of triangle
$\frac{5}{8}=|\frac{1}{2}\times 1\times \frac{1+2\lambda }{2+\lambda }|\Rightarrow \left|\frac{1+2\lambda }{2+\lambda }\right|=\frac{5}{4}\Rightarrow \frac{1+2\lambda }{2+\lambda }=\pm \frac{5}{4}$

Therefore, the equation of $L$ is
$\frac{x}{\frac{5}{4}}+y-1=0\Rightarrow 4x+5y-5=0$

or, $\frac{x}{-\frac{5}{4}}+y-1=0$
$\Rightarrow -4x+5y-5=0\Rightarrow 4x-5y+5=0$