What is the equation of the line that passes through (-3,1) and is perpendicular to the line that passes through the following points: (-2,4),(6,1) ?

Apr 28, 2017

$y = \frac{8}{3} x + 9$

Explanation:

Begin by first finding the slope between $\left(- 2 , 4\right)$ and $\left(6 , 1\right)$ by using the slope formula: $m = \frac{{y}_{2} - {y}_{1}}{{x}_{2} - {x}_{1}}$

If we let $\left(- 2 , 4\right) \to \left(\textcolor{red}{{x}_{1}} , \textcolor{b l u e}{{y}_{1}}\right)$ and $\left(6 , 1\right) \to \left(\textcolor{red}{{x}_{2}} , \textcolor{b l u e}{{y}_{2}}\right)$ then,

$m = \frac{\textcolor{b l u e}{\left(1 - 4\right)}}{\textcolor{red}{\left(6 - \left(- 2\right)\right)}} = - \frac{3}{8}$

Since we are looking for the perpendicular slope we use the following: $m \bot = - \frac{1}{m}$

$m \bot = - \frac{1}{- \frac{3}{8}} = \frac{8}{3}$

Now that we have our slope $\left(\frac{8}{3}\right)$ and given the point $\left(- 3 , 1\right)$ we can find the equation of the line by using the point-slope formula: $y - {y}_{1} = m \left(x - {x}_{1}\right)$

$y - 1 = \frac{8}{3} \left(x - \left(- 3\right)\right) \to y - 1 = \frac{8}{3} \left(x + 3\right)$

We can rewrite this into $y = m x + b$ form if desirable

$y = \frac{8}{3} x + \frac{8}{3} \cdot \frac{3}{1} + 1 \to y = \frac{8}{3} x + \frac{24}{3} + \frac{3}{3} \to y = \frac{8}{3} x + \frac{27}{3}$

Upon further simplification, our final answer is $y = \frac{8}{3} x + 9$