Answer
a) $y=\dfrac{2}{5}x+\dfrac{3}{5}$
b) $2x-5y+3=0$
Work Step by Step
a) The equation of a line in slope-intercept form is:
$$y=mx+b,\tag1$$
where $m$ is the slope and $b$ in the $y$-intercept.
We are given:
$$\begin{align*}
&\text{The line is parallel to the line }2x-5y=10\\
&P(1,1)\text{ belongs to the line}.
\end{align*}$$
Because the line is parallel to the line $2x-5y=10$, it means they have the same slope. We determine that slope:
$$\begin{align*}
2x-5y&=10\\
5y&=2x-10\\
y&=\dfrac{2}{5}x-2\\
m&=\dfrac{2}{5}.
\end{align*}$$
We substitute $m=2/5$ in Eq. $(1)$:
$$y=\dfrac{2}{5}x+b.\tag2$$
Determine $b$ using the coordinates of the point $P$ in Eq. $(2)$:
$$\begin{align*}
1&=\dfrac{2}{5}(1)+b\\
b&=1-\dfrac{2}{5}\\
b&=\dfrac{3}{5}.
\end{align*}$$
The equation of the line is:
$$y=\dfrac{2}{5}x+\dfrac{3}{5}.$$
b) The equation of a line in general form is:
$$Ax+By+C=0.$$
Rewrite the equation $y=(2/5)x+(3/5)$ in general form:
$$\begin{align*}
y&=\dfrac{2}{5}x+\dfrac{3}{5}\\
5y&=2x+3\\
2x-5y+3&=0.
\end{align*}$$
c) Graph the line using the points $\left(0,\dfrac{3}{5}\right)$ and $(1,1)$:
