How do you write an equation of a line passing through (3, -2), perpendicular to #11x-7y=9#?

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Answer 1 · 2016-10-11T17:39:28

Please see the explanation for how it is done.

#y = -7/11x - 1/11#

Begin by writing the given line in slope-intercept form so that you can obtain the slope, m, by observation:

#11x - 7y = 9#

#-11/7x + y = -9/7#

#y = 11/7x - 9/7#

The slope, m, of the given line is:

#m = 11/7#

The slope, n, of all lines perpendicular to the given line is:

#n = -1/m#

#n = -1/(11/7)#

#n = -7/11#

The equation of all lines perpendicular is:

#y = -7/11x + b#

To find the desired line, we substitute the given point, #(3, -2)# for x and y and then solve for b:

#-2 = -7/11(3) + b#

#-22/11 = -21/11 + b#

#-22/11 + 21/11 = b#

#b = -1/11#