Answer
a) $y=-2x$
b) $2x+y=0$
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:
$$\begin{align*}
&\text{The line is perpendicular to the line } y=\dfrac{1}{2}x-10\\
&O(0,0)\text{ belongs to the line}.
\end{align*}$$
Because the line is perpendicular to the line $y=\dfrac{1}{2}x-10$, it means that the product of their slopes is $-1$. We determine that slope $m_1$ of the line $y=(1/2)x-10$:
$$\begin{align*}
m_1&=\dfrac{1}{2}.
\end{align*}$$
Determine the slope $m$ of our line:
$$\begin{align*}
mm_1&=-1\\
m\cdot \dfrac{1}{2}&=-1\\
m&=-2.
\end{align*}$$
We substitute $m=-2$ in Eq. $(1)$:
$$y=-2x+b.\tag2$$
Determine $b$ using the coordinates of the point $O$ in Eq. $(2)$:
$$\begin{align*}
0&=-2(0)+b\\
b&=0.
\end{align*}$$
The equation of the line is:
$$y=-2x.$$
b) The equation of a line in general form is:
$$Ax+By+C=0.$$
Rewrite the equation $y=-2x$ in general form:
$$\begin{align*}
2x+y&=0.
\end{align*}$$
c) Determine another point on the line:
$$\begin{align*}
x&=3\\
y&=-2(3)=-6.
\end{align*}%$$
Graph the line using the points $\left(0,0\right)$ and $(3,-6)$:
