What is the equation of the line that passes through #(-2,1) # and is perpendicular to the line that passes through the following points: #(-16,4),(6,12)?

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Answer 1 · 2016-08-01T17:15:23

Let's first find the equation of the line that it's perpendicular to. We need to find the slope for this:

#m = (y_2 - y_1)/(x_2 - x_1)#

#m = (12 - 4)/(6 - (-16))#

#m = 8/22#

#m = 4/11#

Now, by point-slope form:

#y- y_1 = m(x - x_1)#

#y - 12 = 4/11(x - 6)#

#y - 12= 4/11x - 24/11#

#y = 4/11x - 24/11 + 12#

#y = 4/11x + 108/11#

The slope of a line perpendicular to another always has a slope that is the negative reciprocal of the other line.

Hence, #m_"perpendicular" = -11/4#

Again, by point-slope form:

#y - y_1 = m(x - x_1)#

#y - 1 = -11/4(x - (-2))#

#y - 1 = -11/4x - 11/2#

#y = -11/4x - 11/2 + 1#

#y = -11/4x - 9/2#

#:.#The equation of the line is #y = -11/4x - 9/2#.

Hopefully this helps!