Question

The line $\frac{x}{a}+\frac{y}{b}=1$ meets the $x$-axis at $A$, the $y$-axis at $B$, and the line $y=x$ at $C$ such that the area of $\Delta AOC$ is twice the area of $\Delta BOC$. Then the coordinates of $C$ are

• A. $\left(\frac{b}{3},\frac{b}{3}\right)$
• B. $\left(\frac{2a}{3},\frac{2a}{3}\right)$
• C. $\left(\frac{2b}{3},\frac{2b}{3}\right)$
• D. none of these

Answer 1

Answer: (C)

Given $ar\Delta AOC=2(ar\Delta BOC)$
or $\frac{1}{2}(OA)(x_1)=\frac{2\times 1}{2}(OB)(x_1)$
or $a=2b$
The equation of $AB$ is
$\frac{x}{a}+\frac{y}{b}=1....(i)$
or $\frac{x}{2b}+\frac{y}{b}=1....(ii)$
Since point $C$ lies on line $(ii)$, we have
$\frac{x_1}{2b}+\frac{x_1}{b}=1$
or $x_1=\frac{2b}{3}=\frac{a}{3}$
or $C\equiv (\frac{2b}{3},\frac{2b}{3})$