# How do you write an equation of a line passing through (-6, 1), perpendicular to y = –3x + 1?

Aug 15, 2016

$y = \frac{1}{3} x + 3$

#### Explanation:

Given 2 lines with gradients ${m}_{1} \text{ and } {m}_{2}$. If the lines are perpendicular, then.

$\textcolor{\mathmr{and} a n \ge}{\text{Reminder }} \textcolor{red}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{{m}_{1} \times {m}_{2} = - 1} \textcolor{w h i t e}{\frac{a}{a}} |}}} \ldots \ldots . . \left(A\right)$

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

$\textcolor{red}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{y = m x + b} \textcolor{w h i t e}{\frac{a}{a}} |}}}$
where m represents the gradient and b, the y-intercept.

The equation here is in this form $\Rightarrow m = - 3$

Using (A) The gradient of the line perpendicular to this is

${m}_{\text{perp}} = \frac{- 1}{- 3} = \frac{1}{3}$

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

color(red)(|bar(ul(color(white)(a/a)color(black)(y-y_1=m(x-x_1))color(white)(a/a)|
where $\left({x}_{1} , {y}_{1}\right) \text{ is a point on the line}$

here $m = \frac{1}{3} \text{ and } \left({x}_{1} , {y}_{1}\right) = \left(- 6 , 1\right)$

substitute into point-slope form of the equation.

$y - 1 = \frac{1}{3} \left(x + 6\right) \Rightarrow y - 1 = \frac{1}{3} x + 2$

$\Rightarrow y = \frac{1}{3} x + 3 \text{ is the equation of the line}$