What is the equation of the line in slope-intercept that is perpendicular to the line #4y - 2 = 3x# and passes through the point (6,1)?

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Answer 1 · 2018-03-10T12:10:33

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

Given equation of line is #4y-2=3x#

or, #y=3/4 x +1/2#

Now,for these two lines to be perpendicular product of their slope has to be #-1#

i.e #m(3/4)=-1#

so, #m=-4/3#

Hence,the equation becomes, #y=-4/3x+c#

Given,that this line passes through #(6,1)#,putting the values in our equation we get,

#1=(-4/3)*6 +c#

or, #c=9#

So,the required equation becomes, #y=-4/3 x+9#

or, #3y+4x=27# graph{3y+4x=27 [-10, 10, -5, 5]}