What is the equation of the line in slope-intercept that is perpendicular to the line $4 y - 2 = 3 x$ $4y - 2 = 3x$ 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 $y = m x + c$ $y=mx+c$ where, $m$ $m$ is the slope and $c$ $c$ is the $Y$ $Y$ intercept.

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

or, $y = \frac{3}{4} x + \frac{1}{2}$ $y=3/4 x +1/2$

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

i.e $m (\frac{3}{4}) = - 1$ $m(3/4)=-1$

so, $m = - \frac{4}{3}$ $m=-4/3$

Hence,the equation becomes, $y = - \frac{4}{3} x + c$ $y=-4/3x+c$

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

$1 = (- \frac{4}{3}) \cdot 6 + c$ $1=(-4/3)*6 +c$

or, $c = 9$ $c=9$

So,the required equation becomes, $y = - \frac{4}{3} x + 9$ $y=-4/3 x+9$

or, $3 y + 4 x = 27$ $3y+4x=27$ 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=3x4y - 2 = 3x and passes through the point (6,1)?