How do you write the equation of a line passing through (-6, 2) and perpendicular to y=-3/5x+6?

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Answer 1 · 2018-07-24T12:50:43

#5x-3y+36=0#

The slope of line: #y=-3/5 x+6# is #=-3/5#

Now, the slope #m# of line perpendicular to the given line: #y=-3/5x+6#

#m=-1/{-3/5}#

#=5/3#

Hence the equation of line passing through the point #(x_1, y_1)\equiv(-6, 2)# & having slope #m=5/3# is given by following formula

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

#y-2=5/3(x-(-6))#

#3y-6=5x+30#

#5x-3y+36=0#