Answer
Point-Slope Form: $
y-0=-\dfrac{7}{3}(x-0)
$
Standard Form: $
7x+3y=0
$
Work Step by Step
With the given points, $(0,0)$ and $(3,-7)$, then
\begin{align*}
y_1&=
0
,\\y_2&=
-7
,\\x_1&=
0
,\text{ and }\\ x_2&=
3
.\end{align*}
Using $m=\dfrac{y_1-y_2}{x_1-x_2}$ or the Slope Formula, the slope, $m,$ of the line is
\begin{align*}
m&=\dfrac{0-(-7)}{0-3}
\\\\&=
\dfrac{0+7}{0-3}
\\\\&=
\dfrac{7}{-3}
\\\\&=
-\dfrac{7}{3}
.\end{align*}
The slope, $m,$ is $m=-\dfrac{7}{3}$.
Using $
y-y_1=m(x-x_1)
$ or the Point-Slope form of linear equations, with $m=-\dfrac{7}{3}$ and using one of the given points, $(0,0),$ then
\begin{align*}
y-0&=-\dfrac{7}{3}(x-0)
.\end{align*}
A Point-Slope Form of the given conditions is $
y-0=-\dfrac{7}{3}(x-0)
$.
Using $ax+by=c$ or the Standard Form of linear equations, the equation above is equivalent to
\begin{align*}
y-0&=-\dfrac{7}{3}(x-0)
\\\\
y&=-\dfrac{7}{3}x
\\\\
3(y)&=\left(-\dfrac{7}{3}x\right)3
\\\\
3y&=-7x
\\
3y+7x&=-7x+7x
\\
7x+3y&=0
.\end{align*}
The standard form is $
7x+3y=0
$.
