How do you find the equation of the line that passes through the point (0,2) and perpendicular to the line x-y=6?

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Answer 1 · 2018-03-04T16:22:23

#x+y=2#

let,the equation of the line be #y=mx+c# where, #m# is the slope and #c# is the #Y# intercept.

now,slope of the equation #x-y=6# or #y=x-6# is #1#

We know that for two lines to be mutually perpendicular,their product of slope will have to be #-1#

So,#m*1=-1#

or, #m=-1#

So,the equation becomes, #y=-x+c#

Now,given the line passes through #(0,2)#

So,putting the values we get,

#2=0+c#

So,the equation of the line is, #y=-x+2#

or, #x+y=2# graph{x+y=2 [-10, 10, -5, 5]}