Answer
Point-slope form: $y-3=\dfrac{3}{2}(x+1)$
General form: $3x-2y+9=0$
Work Step by Step
Passing through $(-1,3)$ and parallel to the line whose equation is $3x-2y-5=0$
Rewrite the equation given in slope-intercept form by solving it for $y$:
$3x-2y-5=0$
$3x-5=2y$
$2y=3x-5$
$y=\dfrac{3}{2}x-\dfrac{5}{2}$
Since the slope-intercept form of the equation of a line is $y=mx+b$, the slope of the given equation can be identified as $m=\dfrac{3}{2}$
Since parallel lines have the same slope, the slope of the line whose equation must be found is also $m=\dfrac{3}{2}$
The slope of the line whose equation must be found and a point through which it passes are now known. Substitute them into the point-slope form of the equation of a line formula, which is $y-y_{1}=m(x-x_{1})$
$y-3=\dfrac{3}{2}[x-(-1)]$
$y-3=\dfrac{3}{2}(x+1)$
This is the point-slope form of the equation of the line.
Take all terms to the right side and simplify:
$0=\dfrac{3}{2}(x+1)-y+3$
$0=\dfrac{3}{2}x+\dfrac{3}{2}-y+3$
$0=\dfrac{3}{2}x-y+\dfrac{9}{2}$
Rearrange:
$\dfrac{3}{2}x-y+\dfrac{9}{2}=0$
Multiply the whole equation by $2$:
$2\Big(\dfrac{3}{2}x-y+\dfrac{9}{2}=0\Big)$
$3x-2y+9=0$
This is the general form of the equation of the line
