The line `x/a+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"A O C` is twice the area of `"Delta"B O C` . Then the coordinates of `C` are `(b/3, b/3)` (b) `((2a)/3,(2a)/3)` `((2b)/3,(2b)/3)` (d) none of these

Options

A. `((b)/(3), (b)/(3))`
B. `((2a)/(3), (2a)/(3))`
C. `((2b)/(3), (2b)/(3))`
D. none of these

Answer

C

Explanation

`"Given ar" DeltaAOC=2("ar"DeltaBOC)`
`or (1)/(2)(OA)(x_(1))=(2 xx 1)/(2)(OB)(x_(1))`
or a=2b
The equation of AB is
`(x)/(a)+(y)/(b) = 1 " " (1)`
`"or "(x)/(2b)+(y)/(b) = 1 " " (2)`
Since point C lies on line (2), we have
`(x_(1))/(2b)+(x_(1))/(b) = 1`
`"or "x_(1) = (2b)/(3)=(a)/(3)`
`"or "C-= ((2b)/(3),(2b)/(3))`