# Given that the slope of a line is -1/5, what is the slope of a line that is perpendicular to it?

Jan 28, 2017

Slope is 5

#### Explanation:

The perpendicular to a given slope is its negative reciprocal. This means the fraction is flipped and multiplied by $- 1$. So perpendicular to $- \frac{1}{5}$ is $5$

Jan 28, 2017

$m = 5$

#### Explanation:

The perpendicular slope of any original slope is derived by negating the original slope and then "flipping" the fraction. By "flipping" the fraction, I mean find the inverse of the original slope. So for example:

Original slope: ${m}_{\text{orig}} = - \frac{1}{5}$

Step 1. Negate the original slope. Remember that a negative of a negative is a positive.

$- \left(- \frac{1}{5}\right) = \frac{1}{5}$

Step 2. "Flip" the fraction, finding it's inverse. Remember that whole numbers can be turned automatically into fractions by placing them over a 1.

$\frac{5}{1} = 5 = {m}_{\text{perp}}$

More generally, you can always find the perpendicular slope using this formula:

${m}_{\text{perp")=-1/m_("orig}}$

Jan 28, 2017

All you have to do is remember and follow the method in the first bit of the explanation.

The rest is supportive expansion and contains the actual solution.

#### Explanation:

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Let the slope (gradient) of the first line be $m$

Then the gradient of the perpendicular line is $- \frac{1}{m}$
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$\textcolor{b l u e}{\text{Comment}}$

This s true for any straight or curved line graph.

The only difference is that for a straight line it is a constant value but for a curved line it changes to suit the gradient at each and every point

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$\textcolor{b l u e}{\text{The calculation}}$
The given slope is $- \frac{1}{5}$ which is a constant. Thus the graph is that of a straight line.

The gradient of the perpendicular is: $\left(- 1\right) \times \left(- \frac{5}{1}\right) = + \frac{5}{1}$
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$\textcolor{b l u e}{\text{Foot note}}$

I left the answer in the format of $\frac{5}{1}$ as it represents a ratio.

For every 1 along you go up 5

The teacher will expect you to write the answer gradient as 5 and not $\frac{5}{1}$