# How do you find a standard form equation for the line with (5,4) perpendicular to the line 3x+2y=7?

Sep 15, 2017

$2 x - 3 y = - 2$

#### Explanation:

$3 x + 2 y = 7$ has a slope of $- \frac{3}{2} \textcolor{w h i t e}{\text{xxxx}}$ see Note 1

All lines perpendicular to $3 x + 2 y = 7$ have a slope of $\frac{2}{3} \textcolor{w h i t e}{\text{xxxx}}$see Note 2

If such a perpendicular line goes through $\left(5 , 4\right)$
then we can write it equation in slope point form as:
$y - 4 = \frac{2}{3} \left(x - 5\right) \textcolor{w h i t e}{\text{xxxxxxxxxxxxx}}$see Note 3

This can be converted into standard form as:
2x-3y=-2color(white)("xxxxxxxxxxxxxxxx"see Note 4

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Note 1
A relation in the form $A x + B y = C$ has a slope of $- \frac{A}{B}$;
in this case $A = 3$ and $B = 2$

If you are not familiar with this rule, you can convert given relation $3 x + 2 y = 7$ into slope-intercept form:

$2 y = - 3 x + 7$

$y = \left(- \frac{3}{2}\right) x + \frac{7}{2}$ with slope $\left(- \frac{3}{2}\right)$ and y-intercept $\frac{7}{2}$

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Note 2
If a line has a slope of $m$ then all line perpendicular to it have a slope of $\left(- \frac{1}{m}\right)$

In this case $- \frac{1}{- \frac{3}{2}} = \frac{2}{3}$

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Note 3

A line with slope $m$ through a point $\left({x}_{0} , {y}_{0}\right)$
has a slope-point form:
$\textcolor{w h i t e}{\text{XXX}} y - {y}_{0} = m \left(x - {x}_{0}\right)$

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Note 4
"standard form" for a linear equation is
$\textcolor{w h i t e}{\text{XXX}} A x + B y = C$ with $A , B , C \in \mathbb{Z} , A \ge 0$

converting $y - 4 = \frac{2}{3} \left(x - 5\right)$ into this form:
$\textcolor{w h i t e}{\text{XXX}} 3 \left(y - 4\right) = 2 \left(x - 5\right)$

$\textcolor{w h i t e}{\text{XXX}} 3 y - 12 = 2 x - 10$

$\textcolor{w h i t e}{\text{XXX}} - 2 x + 3 y = 2$

$\textcolor{w h i t e}{\text{XXX}} 2 x - 3 y = - 2$

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