How do you find the slope that is perpendicular to the line #-15 +3y = -12x#?

1 Answer
Nov 23, 2017

Answer:

See a solution process below:

Explanation:

First, we need to find the slope of the line in the problem. We can transform this equation into Standard Linear form. The standard form of a linear equation is: #color(red)(A)x + color(blue)(B)y = color(green)(C)#

Where, if at all possible, #color(red)(A)#, #color(blue)(B)#, and #color(green)(C)#are integers, and A is non-negative, and, A, B, and C have no common factors other than 1

#-15 + color(red)(15) + color(blue)(12x) + 3y = color(blue)(12x) - 12x + color(red)(15)#

#0 + 12x + 3y = 0 + 15#

#12x + 3y = 15#

#(12x + 3y)/color(red)(3) = 15/color(red)(3)#

#(12x)/color(red)(3) + (3y)/color(red)(3) = 5#

#color(red)(4)x + color(blue)(1)y = color(green)(15)#

The slope of an equation in standard form is: #m = -color(red)(A)/color(blue)(B)#

Therefore, the slope of the line in the problem is:

#m = -color(red)(4)/color(blue)(1) = -4#

Let's call the slope of a perpendicular line: #m_p#

The slope of a perpendicular line is the negative inverse of the slope of the line it is perpendicular to.:

#m_p = -1/m#

Substituting gives:

#m_p = -1/-4 = 1/4#