supremetylor29
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PLEASE HELP
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)

Respuesta :

kudzordzifrancis
Part A
The given line passes through (-2,2) and it is parallel to the line

[tex]4x - 3y - 7 = 0[/tex]


We need to determine the slope of this line by writing it in slope -intercept form.


[tex]3y = 4x - 7[/tex]


[tex]y = \frac{4}{3} x - \frac{7}{3} [/tex]
The slope of this line is
[tex] \frac{4}{3} [/tex]


The line parallel to this line also has slope

[tex] m = \frac{4}{3} [/tex]


The equation is

[tex]y = \frac{4}{3} x + c[/tex]


We substitute (-2,2)

[tex]2 = \frac{4}{3}( - 2) + c[/tex]


[tex]c = 2 + \frac{8}{3} = \frac{14}{3} [/tex]


The required equation is


[tex]y = \frac{4}{3} x + \frac{14}{3} [/tex]


PART B


The given line is


[tex]4x - 3y - 7 = 0[/tex]




The slope of this line is
[tex] \frac{4}{3} [/tex]


The slope of the line perpendicular to it is

[tex] m = - \frac{3}{4} [/tex]

The equation of the line is

[tex]y = - \frac{ 3}{4} x + c[/tex]




We substitute the point, (-2,2)



[tex]2= - \frac{ 3}{4} ( - 2) + c[/tex]


[tex]2= \frac{ 3}{2} + c[/tex]



[tex]c = 2 - \frac{3}{2} = \frac{1}{2} [/tex]


The equation of the perpendicular line is


[tex]y= - \frac{ 3}{4} x + \frac{1}{2} [/tex]

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