How do you write an equation of a line passing through (1, -3), perpendicular to # y = 7x + 2#?

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Answer 1 · 2016-09-03T01:39:30

The desired equation is
#x+7y+20=0#

As the equation #y=7x+2# is already in slope intercept form, the slope is #7#.

Now, as the product of slopes of two perpendicular lines is #-1#, the slope of line perpendicular to #y=7x+2# is #-1/7#.

Now the equation of a line passing through #(x_1,y_1)# and having a slope of #m# is

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

Hence equation of a line passing through #(1,-3)# and having a slope of #-1/7# id

#y-(-3)=-1/7×(x-1)# or

#y+3=-1/7×(x-1)# or

#7y+21=-x+1# or

#x+7y+20=0#