Answer
a) $y=-3x+10$
b) $3x+y-10=0$
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:
$$\begin{align*}
&\text{The line is perpendicular to the line } x-3y+16=0\\
&P(1,7)\text{ belongs to the line}.
\end{align*}$$
Because the line is perpendicular to the line $x-3y+16=0$, it means that the product of their slopes is $-1$. We determine that slope $m_1$ of the line $x-3y+16$:
$$\begin{align*}
x-3y+16&=0\\
3y&=x+16\\
y&=\dfrac{1}{3}x+\dfrac{16}{3}\\
m_1&=\dfrac{1}{3}.
\end{align*}$$
Determine the slope $m$ of our line:
$$\begin{align*}
mm_1&=-1\\
m\cdot \dfrac{1}{3}&=-1\\
m&=-3.
\end{align*}$$
We substitute $m=-3$ in Eq. $(1)$:
$$y=-3x+b.\tag2$$
Determine $b$ using the coordinates of the point $P$ in Eq. $(2)$:
$$\begin{align*}
7&=-3(1)+b\\
b&=10.
\end{align*}$$
The equation of the line is:
$$y=-3x+10.$$
b) The equation of a line in general form is:
$$Ax+By+C=0.$$
Rewrite the equation $y=-3x+10$ in general form:
$$\begin{align*}
-3x+10&=y\\
3x+y-10&=0.
\end{align*}$$
c) Determine the $y$-intercept:
$$\begin{align*}
x&=0\\
y&=-3(0)+10=10.
\end{align*}$$
Graph the line using the points $\left(1,7\right)$ and $(0,10)$:
