Answer
(a) $y=3x-16$
(b) $y=-\frac{1}{3}x+4$
Work Step by Step
The given line passes through the points $(2,2)$ and $(1,-1)$.
$\text{Slope of the given line}=\frac{-1-2}{1-2}=\frac{-3}{-1}=3$.
The point given is $(x_{1},y_{1})=(6,2)$.
(a) For parallel line, the slope is same.
Using point-slope form $y-y_{1}=m(x-x_{1})$, we have
$y-2=3(x-6)$
Using distributive property, we get
$y-2=3x-18$
Adding $2$ on both sides, we obtain
$y=3x-16$.
An equation of the line that passes through the given point and is parallel to the given line is $y=3x-16$.
(b) For perpendicular lines, the slopes are negative reciprocals.
$\implies m=-(\frac{1}{3})=-\frac{1}{3}$.
Using point-slope form, we have
$y-2=-\frac{1}{3}(x-6)$
Using distributive property, we get
$y-2=-\frac{1}{3}x+2$
$\implies y-2+2=-\frac{1}{3}x+2+2$
$\implies y=-\frac{1}{3}x+4$
An equation of the line that passes through the given point and is perpendicular to the given line is $y=-\frac{1}{3}x+4$
