Answer
(a) $y=3x-11$
(b) $y=-\frac{1}{3}x-1$
Work Step by Step
The given line passes through the points $(1,-4)$ and $(2,-1)$.
$\text{Slope of the given line}=\frac{-1-(-4)}{2-1}=3$.
The point given is $(x_{1},y_{1})=(3,-2)$
(a) For parallel line, the slope is same.
Using point-slope form $y-y_{1}=m(x-x_{1})$, we have
$y-(-2)=3(x-3)$
Using distributive property, we get
$y+2=3x-9$
$\implies y=3x-9-2$
$\implies y=3x-11$
An equation of the line that passes through the given point and is parallel to the given line is $y=3x-11$.
(b) For perpendicular lines, the slopes are negative reciprocals.
$\implies m=-(\frac{1}{3})=-\frac{1}{3}$
Using point-slope form $y-y_{1}=m(x-x_{1})$, we have
$y-(-2)=-\frac{1}{3}(x-3)$
Using distributive property, we get
$y+2=-\frac{1}{3}x+1$
$\implies y=-\frac{1}{3}x+1-2$
$\implies y=-\frac{1}{3}x-1$
An equation of the line that passes through the given point and is perpendicular to the given line is $y=-\frac{1}{3}x-1$.
