What is the equation in standard form of a perpendicular line to y=3x+6 that passes through (5,-1)?

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What is the equation in standard form of a perpendicular line to y=3x+6 that passes through (5,-1)?

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Answer 1 · 2018-02-08T23:14:08

$y = - \frac{1}{3} x + \frac{2}{3}$ $y=-1/3x+2/3$

Explanation:

first, we need to identify the gradient of the line y=3x+6.
It is already written in the form y=mx+c, where m is the gradient.
the gradient is 3

for any line that is perpendicular, the gradient is $- \frac{1}{m}$ $-1/m$
the gradient of the perpendicular line is $- \frac{1}{3}$ $-1/3$

Using the formula $y - y_{1} = m (x - x_{1})$ $y-y_1=m(x-x_1)$ we can work out the equation of the line.
substitute m with the gradient $- \frac{1}{3}$ $-1/3$
substitute $y_{1}$ $y_1$ and $x_{1}$ $x_1$ with the coordinates given: (5,-1) in this case.
$y - - 1 = - \frac{1}{3} (x - 5)$ $y--1=-1/3(x-5)$
simplify to get the equation:
$y + 1 = - \frac{1}{3} (x - 5)$ $y+1=-1/3(x-5)$
$y = - \frac{1}{3} x + \frac{5}{3} - 1$ $y=-1/3x+5/3-1$

$y = - \frac{1}{3} x + \frac{2}{3}$ $y=-1/3x+2/3$

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What is the equation in standard form of a perpendicular line to y=3x+6 that passes through (5,-1)?

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