Question

Show all work to write the equations of the lines, representing the following conditions, in the form y = mx + b, where m is the slope and b is the y-intercept:
Part A: Passes through (−2, 2) and parallel to 4x − 3y − 7 = 0 (2 points)

Part B: Passes through (−2, 2) and perpendicular to 4x − 3y − 7 = 0 (2 points)

Answer 1

Answer:

Part A) $y=\frac{4}{3}x+\frac{14}{3}$

Part B) $y=-\frac{3}{4}x+\frac{1}{2}$

Step-by-step explanation:

Part A) Passes through (−2, 2) and parallel to 4x − 3y − 7 = 0

we have

$4x-3y-7=0$

Isolate he variable y

$3y=4x-7$

$y=\frac{4}{3}x-\frac{7}{3}$

The slope of the given line is

$m=\frac{4}{3}$

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

$m=\frac{4}{3}$

Find the equation of the line in slope intercept form

$y=mx+b$

we have

$m=\frac{4}{3}$

$point\ (-2,2)$

substitute

$2=\frac{4}{3}(-2)+b$

solve for b

$b=2+\frac{8}{3}$

$b=\frac{14}{3}$

therefore

$y=\frac{4}{3}x+\frac{14}{3}$

Part B) Passes through (−2, 2) and perpendicular to 4x − 3y − 7 = 0

we have

$4x-3y-7=0$

Isolate he variable y

$3y=4x-7$

$y=\frac{4}{3}x-\frac{7}{3}$

The slope of the given line is

$m=\frac{4}{3}$

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

$m=-\frac{3}{4}$

Find the equation of the line in slope intercept form

$y=mx+b$

we have

$m=-\frac{3}{4}$

$point\ (-2,2)$

substitute

$2=-\frac{3}{4}(-2)+b$

solve for b

$b=2-\frac{3}{2}$

$b=\frac{1}{2}$

therefore

$y=-\frac{3}{4}x+\frac{1}{2}$