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

Given  line  

A(a, 0), B(0, b) and O(0, 0)

The point on y=x can be taken in the form

$\begin{array}{l}C\left(\alpha ,\alpha \right)\Rightarrow \frac{\alpha }{a}+\frac{\alpha }{b}=1--\left(1\right)\\ Given\Delta AOC=2\Delta BOC\\ \frac{1}{2}\left|a\alpha -0\right|=2\times \frac{1}{2}\left|0-b\alpha \right|\\ \Rightarrow \left|a\alpha \right|=\left|2b\alpha \right|\Rightarrow a=2b\\ From\left(1\right)\frac{\alpha }{2b}+\frac{\alpha }{b}=1\Rightarrow \frac{3\alpha }{2b}=1\Rightarrow \alpha =\frac{2b}{3}\\ \therefore Thepo\mathrm{int}=\left(\alpha ,\alpha \right)=\left(\frac{2b}{3},\frac{2b}{3}\right)\end{array}$