# How do you write an equation in slope-intercept form for a line with (4, -1) and perpendicular to a line with equation 3x + y = 4?

Apr 14, 2017

$y + 1 = \frac{1}{3} \left(x - 4\right)$

#### Explanation:

A linear equation in the form $A x + B y = C$ has a slope of $- \frac{A}{B}$ [Note 1, below]

The given line $3 x + y = 4$ has a slope of $- \frac{3}{1} = - 3$

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If two lines are perpendicular the slope of one is the negative reciprocal of the other.

Any line perpendicular to $3 x + y = 4$ has a slope of $\textcolor{g r e e n}{+ \frac{1}{3}}$

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A line with a slope of $\textcolor{g r e e n}{m}$ through a point $\left(\textcolor{red}{{x}_{1}} , \textcolor{b l u e}{{y}_{1}}\right)$
can be written in slope-point form as:
$\textcolor{w h i t e}{\text{XXX}} y - \textcolor{b l u e}{{y}_{1}} = \textcolor{g r e e n}{m} \left(x - \textcolor{red}{{x}_{1}}\right)$ [Note 2, below]

A line perpendicular to $3 x + y = 4$ through the point $\left(\textcolor{red}{4} , \textcolor{b l u e}{- 1}\right)$
can be written as:
$\textcolor{w h i t e}{\text{XXX}} y - \textcolor{b l u e}{\left(- 1\right)} = \textcolor{g r e e n}{\frac{1}{3}} \left(x - \textcolor{red}{4}\right)$
which would typically be simplified as
$\textcolor{w h i t e}{\text{XXX}} y + 1 = \frac{1}{3} \left(x - 4\right)$

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Here is a graph with these two equations and the required point for verification:

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Note 1
If you are not familiar with slope of $A x + B y = C$ being $- \frac{A}{B}$
it can be easily derived by converting to slope intercept form:
$\textcolor{w h i t e}{\text{XXX}} A x + B y = C$
$\textcolor{w h i t e}{\text{XXX}} \rightarrow B y = - A x + C$
$\textcolor{w h i t e}{\text{XXX}} \rightarrow y = - \frac{A}{B} x + \frac{C}{B}$ with slope of $- \frac{A}{B}$ and y-intercept $\frac{C}{B}$

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Note 2
For some the more common general slope-point form might be
$\textcolor{w h i t e}{\text{XXX}} \frac{\left(y - {\textcolor{b l u e}{y}}_{1}\right)}{x - \textcolor{red}{\left({x}_{1}\right)}} = \textcolor{g r e e n}{m}$
I prefer the form that I have used above since it avoids the exclusion problem when $x = \textcolor{red}{{x}_{1}}$