Question

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

Answer 1

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 `-1/5` is `5`

Answer 2

`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_("orig")=-1/5`

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

`-(-1/5)=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.

`5/1=5=m_("perp")`

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

`m_("perp")=-1/m_("orig")`

Answer 3

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:

`color(blue)("THE REALLY IMPORTANT BIT: The basic rule")`

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

Then the gradient of the perpendicular line is `-1/m`
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`color(blue)("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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`color(blue)("The calculation")`
The given slope is `-1/5` which is a constant. Thus the graph is that of a straight line.

The gradient of the perpendicular is: `(-1)xx(-5/1) = +5/1`
`" "color(red)(uarr)`
`" "color(red)("Inverting the given gradient")`
.................................................................................
`color(blue)("Foot note")`

I left the answer in the format of `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 `5/1`