How do you write the standard form of the equation of the line that is perpendicular to y=-4/3x-7 and contains (-5,-7)?

1 Answer
Jun 27, 2018

The standard form equation is #3x - 4y = 13#.

Explanation:

The slope of our line is #color(red)(-\frac{4}{3})#. The slope of any perpendicular line is given by the negative reciprocal.

#m' = -\frac{1}{m}#

#m' = \frac{1}{color(red)(-\frac{4}{3})}#

#m' = \frac{3}{4}#

The second condition is that the line contains #(color(blue)(-5), color(green)(-7))#.

#y = 3/4x + b#

Substitute #color(blue)(x = -5)#, #color(green)(y = -7)#.

#color(green)(-7) = 3/4(color(blue)(-5)) + b#

#color(green)(-7) = -15/4 + b#

#-13/4 = b#

So in slope-intercept form, the equation of our line is

#y = 3/4x - 13/4#

but the question requires it in standard form. This means that it should be in the form

#Ax + By = C#

where #A#, #B#, and #C# are all integers and #A# is positive. In our equation, we will multiply by the lowest common denominator to eliminate the denominator.

#4y = 3x - 13#

#<=> 3x - 13 = 4y#

#<=> 3x - 4y = 13#