Answer
Point-slope form: $y-2=\dfrac{2}{3}(x+2)$
General form: $2x-3y+10=0$
Work Step by Step
Passing through $(-2,2)$ and parallel to the line whose equation is $2x-3y-7=0$
Rewrite the given line's equation in slope-intercept form. Start by taking $3y$ to the right side:
$2x-3y-7=0$
$2x-7=3y$
Rearrange:
$3y=2x-7$
Take the $3$ to divide the right side:
$y=\dfrac{2}{3}x-\dfrac{7}{3}$
The slope-intercept form of the equation of a line is $y=mx+b$, where $m$ is its slope and $b$ is its $y$-intercept. For the equation given, $m=\dfrac{2}{3}$
Since parallel lines have the same slope, the slope of the line whose equation must be found is also $m=\dfrac{2}{3}$
The slope of the line whose equation must be found and a point through which it passes are now known. Substitute these elements into the point-slope form of the equation of a line formula, which is $y-y_{1}=m(x-x_{1})$ and simplify:
$y-2=\dfrac{2}{3}[x-(-2)]$
$y-2=\dfrac{2}{3}(x+2)$
This is the equation of the line in point-slope form.
Take all terms to the right side:
$0=\dfrac{2}{3}(x+2)-y+2$
Simplify:
$0=\dfrac{2}{3}x+\dfrac{4}{3}-y+2$
$0=\dfrac{2}{3}x-y+\dfrac{10}{3}$
Multiply the whole equation by $3$ and rearrange:
$3\Big(0=\dfrac{2}{3}x-y+\dfrac{10}{3}\Big)$
$0=2x-3y+10$
$2x-3y+10=0$
This is the equation of the line in general form.
