Answer
$\text{(a)}$
\begin{align}
A = \frac{4}{3}
\end{align}
$\text{(b)}$
\begin{align}
y = 0.413x
\end{align}
Work Step by Step
$\text{(a) First, we have to find the intersection:}$
\begin{align}
& 2x-x^2 = 0 \\
& x = 0 \ and \ x = 2 \\
& A = \int_0^2 (2x-x^2) \ dx = \left[ x^2-\frac{x^3}{3}\right]_0^2 = \frac{4}{3}
\end{align}
$\text{(b) The half of the area in part (a) is $\frac{2}{3}$.}$
$\text{The intersection between y = mx and y = 2x - x$^2$:}$
\begin{align}
& 2x-x^2 = mx \\
& x = 0 \ and \ x = 2-m \\
& \int_0^{2-m} mx \ dx + \int_{2-m}^2 (2x-x^2) \ dx = \left[m\frac{x^2}{2} \right]_0^{2-m} + \left[x^2 -\frac{x^3}{3}\right]_{2-m}^{2} = \\ & = m\frac{4-4m+m^2}{2} -m^2+4m - \frac{m^3-6m^2+12m}{3} = 2m-m^2+\frac{m^3}{6} = \frac{2}{3} \\
& \Rrightarrow m^3-6m^2+12m-4=0 \Rrightarrow m = 0.413
\end{align}
$\text{Thus, the line that divides the area in two parts is y = 0.413x}$
