How do you write the equation of the line parallel to the line x + 4y = 6 and passes through (-8, 5)?

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How do you write the equation of the line parallel to the line x + 4y = 6 and passes through (-8, 5)?

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Answer 1 · 2016-04-24T00:33:16

I found $x + 4 y = 12$ $x+4y=12$

Explanation:

We can use the general relationship for the equation of a line passing through $(x_{0}, y_{0})$ $(x_0,y_0)$ and slope $m$ $m$ as:
$y - y_{0} = m (x - x_{0})$ $y-y_0=m(x-x_0)$
to be parallel the slope must be the same of your original line.
We write the original line (collecting $y$ $y$ ) as:
$y = - \frac{1}{4} x + \frac{6}{4}$ $y=-1/4x+6/4$
whose slope is $- \frac{1}{4}$ $-1/4$
so we can find our parallel as:
$y - 5 = - \frac{1}{4} [x - (- 8)]$ $y-5=-1/4[x-(-8)]$
$4 y - 20 = - x - 8$ $4y-20=-x-8$
so the equation will be:
$x + 4 y = 12$ $x+4y=12$

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How do you write the equation of the line parallel to the line x + 4y = 6 and passes through (-8, 5)?

1 Answer

Explanation:

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