How do you write an equation for the line in slope intercept form that is perpendicular to the given line and that passes through the given point –4x + 12y = 18; (–2, 4)?

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Answer 1 · 2018-07-03T04:36:59

#y=-3x-2#

The slope of straight line #-4x+12y=18# or #y=1/3x+9/2# is
#1/3#

The unknown line is perpendicular to the given line: #-4x+12y=18# hence, the slope of unknown line

#=-1/(1/3)=-3#

Now, the equation of line with a slope #m=-3# passing through the point #(-2, 4)\equiv(x_1, y_1)# is given as follows

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

#y-4=-3(x-(-2))#

#y=-3x-2#

Above is the slope intercept form of line