Answer:
Part A) [tex]y=\frac{4}{3}x+\frac{14}{3}[/tex]
Part B) [tex]y=-\frac{3}{4}x+\frac{1}{2}[/tex]
Step-by-step explanation:
Part A) Passes through (−2, 2) and parallel to 4x − 3y − 7 = 0
we have
[tex]4x-3y-7=0[/tex]
Isolate he variable y
[tex]3y=4x-7[/tex]
[tex]y=\frac{4}{3}x-\frac{7}{3}[/tex]
The slope of the given line is
[tex]m=\frac{4}{3}[/tex]
Remember that
If two lines are parallel then their slopes are the same
therefore
The slope of the line parallel to the given line is also
[tex]m=\frac{4}{3}[/tex]
Find the equation of the line in slope intercept form
[tex]y=mx+b[/tex]
we have
[tex]m=\frac{4}{3}[/tex]
[tex]point\ (-2,2)[/tex]
substitute
[tex]2=\frac{4}{3}(-2)+b[/tex]
solve for b
[tex]b=2+\frac{8}{3}[/tex]
[tex]b=\frac{14}{3}[/tex]
therefore
[tex]y=\frac{4}{3}x+\frac{14}{3}[/tex]
Part B) Passes through (−2, 2) and perpendicular to 4x − 3y − 7 = 0
we have
[tex]4x-3y-7=0[/tex]
Isolate he variable y
[tex]3y=4x-7[/tex]
[tex]y=\frac{4}{3}x-\frac{7}{3}[/tex]
The slope of the given line is
[tex]m=\frac{4}{3}[/tex]
Remember that
If two lines are perpendicular, then their slopes are opposite reciprocal
therefore
The slope of the line perpendicular to the given line is
[tex]m=-\frac{3}{4}[/tex]
Find the equation of the line in slope intercept form
[tex]y=mx+b[/tex]
we have
[tex]m=-\frac{3}{4}[/tex]
[tex]point\ (-2,2)[/tex]
substitute
[tex]2=-\frac{3}{4}(-2)+b[/tex]
solve for b
[tex]b=2-\frac{3}{2}[/tex]
[tex]b=\frac{1}{2}[/tex]
therefore
[tex]y=-\frac{3}{4}x+\frac{1}{2}[/tex]
