Question 905545
 write an equation that is part one: parallel and then also part two: perpendicular to the equation y=2/3x-4 passing through the point (-2,-5). Write your answer in standard form. 
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y = {{{2/3}}}x - 4
equations of parallel lines have the same slope
Using the slope intercept form y = mx + b
Replace x and y with the given point, x=-2; then y = -5, slope is 2/3
-5 = {{{2/3}}}(-2) + b
-5 = {{{-4/3}}} + b
-5 + {{{4/3}}} = b
{{{-15/3}}} + {{{4/3}}} = b
{{{-11/3}}} = b
the equation
y = {{{2/3}}}x - {{{11/3}}}
The standard form ax + by = c, with all integer coefficients
Multiply our equation by 3 to get rid of the denominators
3y = 2x - 11
-2x + 3y = -11
or you can mult by -1 and you have fewer negatives
2x - 3y = 11
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perpendicular, the slopes relationship; m1*m2 = -1
Assuming m1 = 2/3, find m2
{{{2/3}}} * m2 = -1
Multiply both sides by 3/2 and you have
m2 = {{{-3/2}}}Using the slope intercept form y = mx + b
Replace x and y with the given point, x=-2; then y = -5, slope is -3/2
-5 = {{{-3/2}}}(-2) + b
-5 = 3 + b
-5 - 3 = b
b = -8
the equation
y = {{{-3/2}}}x - 8
The standard form ax + by = c, with all integer coefficients
Multiply our equation by 2 to get rid of the denominators
2y = -3x - 16
3x + 2y = -16 is the standard form
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You can check this for yourself replace x with -2 and y with -5, equals -16