The idea of **slope** is very important not only in mathematics on the **plane**, but in 3-D math and math in higher dimensions (yup!). We need a way to say *how* sloped a graph is, and in which *direction* it's sloping.

If you're a skier, you know there are slopes that slope straight down the mountain, slopes that slope both down and off to the side (double slopes or double fall lines), gentle slopes and steep slopes. In higher math, slope is called **gradient**.

It's a concept vital for understanding the behavior of things like electric and magnetic fields, weather phenomena, heat transfer, &c.

In the plane, we define slope very simply as the amount of rise (or fall) of the dependent variable divided by the change in the independent variable over the same interval: **rise over run**.

**Slope** is rise divided by run ("**rise over run**").

$$\text{slope} = \frac{\Delta y}{\Delta x} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{\text{rise}}{\text{run}}$$

The slope of a line is calculated by first finding two points (any two points will do) that lie on the line (← red dots at left).

Construct the right triangle with sides **Δx = (x _{2} - x_{1})** and

The slope of a horizontal line has a zero numerator so its value is zero.

The slope of a vertical line has a zero denominator. Division by zero gives an infinite result. Think about it: as the denominator of a fraction decreases, the value of the fraction increases. If we continue that trend until the denominator vanishes, we get an infinitely large fraction.

Calculate the slope of the line containing points (11, -4) and (-5, 23).

**Solution**

$$ \begin{align} m &= \frac{y_2 - y_1}{x_2 - x_1} \\[5pt] &= \frac{23 - (-4)}{-5 - 11} = -\frac{27}{16} \end{align}$$

As you work through some of the problems below, you'll notice that lines can have positive and negative slope (as well as zero and infinite slope). A line with positive slope is lower on the left (-x) side than the right (+x), and a line with negative slope is the opposite: up on the left, down on the right.

Normally in mathematics, we just express slope as a fractional or decimal quantity. A slope of 1 means that for every meter of run, there is a meter of rise (and the slope angle is 45˚ — think about it - it's an isosceles triangle).

On the roadway you'll see hills measured in percent grade. Percent grade is determined by measuring the number of feet (any unit will do) of rise for every 100 feet of horizontal A distance (run).

On the road, an 8% grade is a pretty steep hill - steep enough that gearing down is a good idea. But if you work out the angle of the hill, it's only about 4.5˚. Cars just roll very easily!

Watch out for trucks parked on triangles!

1. |
## Solution$$m = \frac{2 - 6}{2 - (-1)} = -\frac{4}{3}$$ |

2. |
## Solution$$m = \frac{3 - 2}{9 - 1} = \frac{1}{8}$$ |

3. |
## Solution$$m = \frac{-4 - (-4)}{3 - (-4)} = \frac{0}{7}$$ Slope = 0, a horizontal line. |

4. |
## Solution$$m = \frac{8 - 1}{11 - 6} = \frac{7}{5}$$ |

5. |
## Solution$$m = \frac{2 + 5}{-1 + 3} = \frac{7}{2}$$ |

6. |
## Solution$$m = \frac{-4 - 20}{6 - 6} = \frac{-24}{0} \rightarrow \infty$$ Infinite slope, a vertical line. |

The slopes of any two parallel lines must be identical. If the slope of a one line is different from that of another, the two lines will eventually intersect somewhere on the plane.

The slopes of two lines that are perpendicular (intersect at right angles) are **negative reciprocals** of one another. Note: sometimes students refer to this as "opposite reciprocal", but "opposite" doesn't have any real meaning in mathematics. Here are some slopes and their negative reciprocals:

Slope of line |
Slope of ⊥ line |
---|---|

-½ | 2 |

0.1 | -10 |

0 | ∞ |

-5 | ^{1}/_{5} |

The slopes of **parallel** lines are identical, and lines with identical slopes are parallel (and they could be the *same* line).

The slopes of **perpendicular** lines are negative reciprocals of each-another, and lines with negative-reciprocal slopes are perpendicular and must intersect somewhere.

7. |
Find the slope of a line perpendicular to the line passing through (5, 3) and (1, -12). ## Solution$$ \begin{align} m &= \frac{-12 - 3}{1 - 5} = \frac{15}{4} \\[5pt] \text{so} \: \: m_{\perp} &= -\frac{4}{15} \end{align}$$ |

8. |
Find the slope of a line perpendicular to the line passing through (-3, -2) and (1, 6). ## Solution$$ \begin{align} m &= \frac{6 + 2}{1 + 3} = \frac{8}{4} = 2 \\[5pt] \text{so} \: \: m_{\perp} &= -\frac{1}{2} \end{align}$$ |

Curves have slope, too – it's just a *changing* slope. The slope of a curve at any point on it is the slope of the line **tangent** to (*just* touching) the curve at that point.

The slope of a curve at a maximum or minimum point is zero.^{}

The slope of a curve can be calculated using differential calculus, and it's not all that mysterious. You'll get there; just keep on doing what you're doing!

Given the two parallel lines **L** and **M**, draw the transversal **T** parallel to the y-axis. Then draw segments b and d as shown, parallel to the x-axis, forming the right angles shown. Because **L||M, ∠ 1 ≅ ∠ 2** (corresponding angles).

That means the two triangles are similar, therefore the ratios of their corresponding sides are equal. In particular, ^{a}**/ _{b} = ^{c}/_{d}**. Because these are also the slopes of lines

Let lines **L** and **M** be perpendicular. Draw **T, V,** and **a** and **c** parallel to the coordinate axes, so that the two right triangles shown are formed. Let **b ≅ d**. Now **∠ 1 ≅ ∠ 2** by subtraction of the intervening angle (3) from 90˚ angles, so the two right triangles are congruent by **ASA**.

That means **-a = c** and **b = d**, so the slope of line **M** is ^{c}**/ _{d} **and the slope of

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