Question

$L_1$ is a line intersecting $x$ and $y$ axes at $A(a,0)$ and $B(0,b).$ $L_2$ is a line perpendicular to $L_1$ intersecting $x$ and $y$ axes at $C$ and $D$ respectiveley. If the area of the triangle $OCD$ is 4 times the area of triangle $OAB$, then equation of $AD$ is

• A. $x+2y=26$
• B. $2x+y+2a=0$
• C. $x-2y+2b=0$
• D. $2x-y-2a=0$

Answer 1

The correct option is D $2x-y-2a=0$

Equation of $AB:\frac{x}{a}+\frac{y}{b}=1$
$L_2\perp L_1\Rightarrow \frac{x}{b}-\frac{y}{a}=\alpha$
$\Rightarrow \frac{x}{\alpha b}+\frac{y}{-\alpha a}=1$
$C:(\alpha b,0),D:(0,-\alpha a)$
$Ar\left(\Delta OCD\right)$
$=|\frac{1}{2}\times OC\times OD|=|\frac{1}{2}\times (\alpha b\left)\right(-\alpha a\left)\right|=\frac{{\alpha }^2}{2}|ab|$

$Ar\left(\Delta OAB\right)$
$=|\frac{1}{2}\times OA\times OB|=\frac{1}{2}|ab|$
$\Rightarrow \frac{{\alpha }^2}{2}|ab|=4\times \frac{1}{2}|ab|\Rightarrow \alpha =\pm 2$
$A(a,0),D(0,\mp 2a)$
Line passing through $A$ and $D$
$\Rightarrow \frac{x}{a}+\frac{y}{\mp 2a}=1\Rightarrow 2x\mp y-2a=0$