Question

Find the equation of the line passing through the origin and dividing the curvilinear triangle with vertex at the origin, bounded by the curves $y=2x-x^2,y=0$ and $x=1$ into two parts of equal area.

• A. $y=2x/3$
• B. $y=x/3$
• C. $y=2x/5$
• D. $y=5x/3$

Answer 1

The correct option is A $y=2x/3$

$-y=(x-1{)}^2-1$
$y=1-(x-1{)}^2$
Hence the required area
${\int }_0^{1}1-(x-1{)}^2.dx$
$=[x-\frac{(x-1{)}^3}{3}{]}_0^1$
$=1-(-\frac{(-1{)}^3}{3})$
$=1-\frac{1}{3}$
$=\frac{2}{3}$.
Hence $y=\frac{2x}{3}$ will divide the entire triangle in 2 equal parts.