# How do you find the equation of the line passing through (3, -1) and perpendicular to 2x+7y=-1?

Apr 29, 2017

$y = \frac{7}{2} x - \frac{23}{2}$

#### Explanation:

The equation of a line in $\textcolor{b l u e}{\text{point-slope form}}$ is.

$\textcolor{red}{\overline{\underline{| \textcolor{w h i t e}{\frac{2}{2}} \textcolor{b l a c k}{y - {y}_{1} = m \left(x - {x}_{1}\right)} \textcolor{w h i t e}{\frac{2}{2}} |}}}$
where m represents the slope and $\left({x}_{1} , {y}_{1}\right) \text{ a point on the line}$

$\text{Require to know the following fact}$

$\text{For 2 perpendicular lines with slopes " m_1" and } {m}_{2}$

$\text{then } \textcolor{red}{\overline{\underline{| \textcolor{w h i t e}{\frac{2}{2}} \textcolor{b l a c k}{{m}_{1} \times {m}_{2} = - 1} \textcolor{w h i t e}{\frac{2}{2}} |}}}$

$\text{rearrange " 2x+7y=-1" into slope-intercept form}$

$\text{that is " y=mx+b" where m is the slope}$

$\text{subtract 2x from both sides}$

$\cancel{2 x} \cancel{- 2 x} + 7 y = - 2 x - 1$

$\Rightarrow 7 y = - 2 x - 1$

$\text{divide ALL terms by 7}$

$\frac{\cancel{7} y}{\cancel{7}} = - \frac{2}{7} x - \frac{1}{7}$

$\Rightarrow y = - \frac{2}{7} x - \frac{1}{7} \leftarrow \textcolor{red}{\text{ in form y=mx+b}}$

$\Rightarrow \text{slope } = m = - \frac{2}{7}$

$\Rightarrow {m}_{\text{perpendicular}} = \frac{- 1}{- \frac{2}{7}} = \frac{7}{2}$

$\text{using " m=7/2" and } \left({x}_{1} , {y}_{1}\right) = \left(3 , - 1\right)$

$y - \left(- 1\right) = \frac{7}{2} \left(x - 3\right)$

$\Rightarrow y + 1 = \frac{7}{2} \left(x - 3\right) \leftarrow \textcolor{red}{\text{ in point-slope form}}$

$\text{distribute and simplify}$

$y = \frac{7}{2} x - \frac{21}{2} - 1$

$\Rightarrow y = \frac{7}{2} x - \frac{23}{2} \leftarrow \textcolor{red}{\text{ in slope-intercept form}}$