question_answer

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

Correct Answer: C

Solution :

[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 \[C\equiv \left( \frac{2b}{3},\frac{2b}{3} \right)\]