# What is the slope of any line perpendicular to the line passing through (3,6) and (-8,4)?

Feb 24, 2016

$- \frac{11}{2}$

#### Explanation:

$\textcolor{m a \ge n t a}{\text{Introduction to how it works}}$

Standard form of the equation of a straight line is: $y = m x + c$
Where $m$ is the gradient (slope)

$\textcolor{g r e e n}{\text{Any line perpendicular to the original line has the slope of: }}$
$\textcolor{g r e e n}{\left(- 1\right) \times \frac{1}{m}}$

So for the second line the equation changes

$\textcolor{b l u e}{\text{From ")color(brown)(y=mx+c )color(blue)(" to }} \textcolor{g r e e n}{y = - \frac{1}{m} x + c}$

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$\textcolor{m a \ge n t a}{\text{Answering your question}}$

$\textcolor{b l u e}{\text{Determine the gradient of the given line}}$
Let the first listed coordinates be the first point

$\left({x}_{1} , {y}_{1}\right) \to \left(3 , 6\right)$
$\left({x}_{2} , {y}_{2}\right) \to \left(- 8 , 4\right)$

Gradient given line$\to \left(\text{change in y-axis")/("change in x-axis left to right}\right)$

Gradient given line (m)$\to \frac{{y}_{2} - {y}_{1}}{{x}_{2} - {x}_{1}} \to \frac{4 - 6}{\left(- 8\right) - 3} \to \frac{- 2}{- 11} = + \frac{2}{11}$

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$\textcolor{b l u e}{\text{Determine the gradient of the line perpendicular to the first one}}$

$\left(- 1\right) \times \frac{1}{m} = - \frac{11}{2}$