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]}

What is the equation of the line in slope-intercept that is perpendicular to the line 4y - 2 = 3x4y - 2 = 3x and passes through the point (6,1)?