Question

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

Answer 1

The point of intersection of ( $\frac{x}{1}+\frac{y}{b}$ ) = 1

And y=x would be ($\frac{1\times b}{(1+b)},\frac{1\times b}{(1+b)}$).

The area of triangle AOC (0,0), (1,0) and ($\frac{1\times b}{(1+b)},\frac{1\times b}{(1+b)}$), would be $\frac{1}{2}\times \frac{1^{2}b}{(1+b)}$

The area of triangle BOC (0,0), (0,b) and ($\frac{1\times b}{(1+b)},\frac{1\times b}{(1+b)}$)., would be $2\times \frac{1}{2}\times \frac{1^{2}b}{(1+b)}$

Since, area of $\Delta AOC$ is twice the area of $\Delta BOC$

Therefore,

$\frac{1}{2}\times \frac{1^{2}b}{(1+b)}$ =$2\times \frac{1}{2}\times \frac{1^{2}b}{(1+b)}$

$⟹1=2b.$

coordinate of C=$(\frac{1b}{(1+b)},\frac{1b}{(1+b)})$.

Therefore coordinate of C is $(\frac{1}{3},\frac{1}{3})$.