Answer
(a) An equation of the line that passes through the given point and is parallel to the given line is $y=-4x+19$.
(b) An equation of the line that passes through the given point and is perpendicular to the given line is $y=\frac{1}{4}x+2$.
Work Step by Step
The given line passes through the points $(1,6)$ and $(2,2)$.
$\text{Slope of the given line}=\frac{2-6}{2-1}=-4$.
The point given is $(x_{1},y_{1})=(4,3)$
(a) For parallel line, the slope is same.
Using point-slope form $y-y_{1}=m(x-x_{1})$, we have
$y-3=-4(x-4)$
Using distributive property, we get
$y-3=-4x+16$
$\implies y=-4x+16+3$
$\implies y=-4x+19$
An equation of the line that passes through the given point and is parallel to the given line is $y=-4x+19$.
(b) For perpendicular lines, the slopes are negative reciprocals.
$\implies m=-(\frac{1}{-4})=\frac{1}{4}$
Using point-slope form $y-y_{1}=m(x-x_{1})$, we have
$y-3=\frac{1}{4}(x-4)$
Using distributive property, we get
$y-3=\frac{1}{4}x-1$
$\implies y=\frac{1}{4}x-1+3$
$\implies y=\frac{1}{4}x+2$
An equation of the line that passes through the given point and is perpendicular to the given line is $y=\frac{1}{4}x+2$.
