Slope and the Slope Criteria | Geometry | LevelKite

You already know how to name the coordinates of a point. Now you can use coordinates to measure a line. Slope is a number that tells you how steep a line is and which way it tilts. (One kind of line has no slope number; you will meet it soon.)

To see where the number comes from, split a trip along a line into two moves: the change in $x$ , how far you move along the horizontal (left-right) $x$ -axis, and the change in $y$ , how far you move along the vertical (up-down) $y$ -axis.

In the figure below, a line passes through Point A at $(1, 1)$ and Point B at $(5, 3)$ . Follow the horizontal guide line from Point A to Point C at $(5, 1)$ : $x$ goes from $1$ to $5$ , a change of $4$ . Then follow the vertical guide line up to Point B: $y$ goes from $1$ to $3$ , a change of $2$ . The line climbs $2$ units for every $4$ units across, which is $\frac{2}{4} = \frac{1}{2}$ of a unit up for each unit to the right. That rate is the slope.

From Point A at

(1, 1)

to Point C at

(5, 1)

the change in

x

is

4

; from Point C to Point B at

(5, 3)

the change in

y

is

2

.

Here is the exact definition. A line is vertical if all of its points share one $x$ -coordinate, and horizontal if all of its points share one $y$ -coordinate. The slope $m$ of a nonvertical line is the change in $y$ divided by the change in $x$ between two different points on the line. Any two different points on the line give the same value, so use the ones you are given.

Subtract in the same order on the top and the bottom, so both changes describe the same trip. A positive slope means the line rises as you move right, and a negative slope means it falls. For points $(x_1, y_1)$ and $(x_2, y_2)$ , where the small numbers just label the first and second point, the definition reads as below. The change in $x$ cannot be $0$ , because division by zero is undefined.

m = \frac{y_2 - y_1}{x_2 - x_1}, \qquad x_2 - x_1 \neq 0

Worked example

Slope from two points

Find the slope of the line through $(-2, 5)$ and $(4, 1)$ .

Name the points in the order you will use them. Their x-coordinates are different, so the change in x is not 0 and the definition of slope applies.
$(x_1, y_1) = (-2, 5), \quad (x_2, y_2) = (4, 1)$
Substitute into the definition of slope. The second point goes first on both the top and the bottom, so both changes describe the same trip.
$m = \frac{1 - 5}{4 - (-2)}$
Do each subtraction. Subtracting -2 is the same as adding 2, so the bottom becomes 4 + 2.
$m = \frac{-4}{4 + 2} = \frac{-4}{6}$
Simplify the fraction exactly by dividing the top and the bottom by their common factor, 2.
$m = \frac{-4 \div 2}{6 \div 2} = \frac{-2}{3} = -\frac{2}{3}$
Check by reversing the order of the points. The trip now runs the other way, so both changes switch sign and the quotient stays the same.
$m = \frac{5 - 1}{-2 - 4} = \frac{4}{-6} = -\frac{2}{3}$

Answer

The slope of the line through $(-2, 5)$ and $(4, 1)$ is $-\frac{2}{3}$ , so the line falls $2$ units for every $3$ units it moves to the right.

Two kinds of lines need special care, and the definition of slope handles both.

Take the horizontal line through $(0, 3)$ and $(4, 3)$ . Its change in $y$ is $3 - 3 = 0$ and its change in $x$ is $4 - 0 = 4$ , so its slope is $\frac{0}{4} = 0$ . Every horizontal line has slope zero: it neither rises nor falls.

Now take the vertical line through $(2, 1)$ and $(2, 5)$ . Its change in $x$ is $2 - 2 = 0$ , so the formula would have to divide $5 - 1 = 4$ by $0$ . Division by zero is undefined, so every vertical line has undefined slope. Undefined is not the same as $0$ : slope $0$ is a number, and a vertical line has no slope number at all.

The figure below shows both lines, with the $x$ -axis horizontal and the $y$ -axis vertical. Line A is the vertical line and Line B is the horizontal line, and the marked points are the ones you just used.

Line A has undefined slope because both of its marked points have

x = 2

; Line B has slope

0

because both of its marked points have

y = 3

.

Slopes also tell you how two lines are related. Parallel lines are different lines in the same plane that never meet. Perpendicular lines are lines that meet at a right angle, an angle of $90^\circ$ .

You cannot decide either one by looking at a drawing, because a picture can mislead, so you always work from the coordinates. In the next figure, the $x$ -axis is horizontal and the $y$ -axis is vertical. You are given that Line A passes through $(0, 1)$ and $(1, 3)$ , Line B passes through $(2, 0)$ and $(3, 2)$ , and Line C passes through $(1, 3)$ and $(3, 2)$ .

Using the marked points, Line A and Line B each have slope

2

, and Line C has slope

-\frac{1}{2}

.

Apply the definition of slope to the marked points:

Line A: $\frac{3 - 1}{1 - 0} = \frac{2}{1} = 2$
Line B: $\frac{2 - 0}{3 - 2} = \frac{2}{1} = 2$
Line C: $\frac{2 - 3}{3 - 1} = \frac{-1}{2} = -\frac{1}{2}$

Lines A and B have equal slopes, so each time $x$ grows by $1$ , both lines climb $2$ . The vertical gap between them never changes, so they never meet. They are also different lines: if $(3, 2)$ were on Line A, the slope from $(1, 3)$ to $(3, 2)$ would be $2$ , but it is $-\frac{1}{2}$ .

Now compare Lines A and C. A trip along Line A from $(0, 1)$ to $(1, 3)$ has a change in $x$ of $1$ and a change in $y$ of $2$ . A trip along Line C from $(1, 3)$ to $(3, 2)$ has a change in $x$ of $2$ and a change in $y$ of $-1$ . The numbers $1$ and $2$ traded places, and one sign changed. That is exactly what a quarter turn does to a trip, so these lines meet at a right angle, and their slopes multiply to $2 \cdot \left(-\frac{1}{2}\right) = -1$ .

Key concept

The slope criteria

A criterion (plural criteria) is a test that settles a question. Take two different nonvertical lines with slopes $m_1$ and $m_2$ .

Parallel Lines Slope Criterion: the lines are parallel exactly when $m_1 = m_2$ .
Perpendicular Lines Slope Criterion: the lines are perpendicular exactly when $m_1 m_2 = -1$ .

A vertical line has undefined slope, so neither test applies to it. A vertical line and a horizontal line are perpendicular without a product, because the coordinate plane is built with perpendicular axes and these lines run in the axes' directions.

\text{Different nonvertical lines: } m_1 = m_2 \iff \text{parallel}, \quad m_1 m_2 = -1 \iff \text{perpendicular}

The words exactly when pack a theorem (a statement that can be proved) and its converse into one sentence. A converse switches the if part and the then part. For two different nonvertical lines:

Lines to slopes: if the lines are parallel, then $m_1 = m_2$ ; if the lines are perpendicular, then $m_1 m_2 = -1$ .
Slopes to lines: if $m_1 = m_2$ , then the lines are parallel; if $m_1 m_2 = -1$ , then the lines are perpendicular.

A converse is not automatically true, but here both directions hold, and they do different jobs. Use slopes to lines to conclude that a relationship holds. Use lines to slopes to rule one out: if the slopes differ, the lines cannot be parallel, and if the product is not $-1$ , they cannot be perpendicular. In this lesson you use the criteria as stated, and you will prove them in the next unit.

Worked example

Identifying lines from their slopes

Given: four different nonvertical lines have these slopes. Line P has slope $\frac{3}{4}$ , Line Q has slope $\frac{3}{4}$ , Line R has slope $-\frac{4}{3}$ , and Line S has slope $\frac{4}{3}$ . Decide which pairs are parallel, which are perpendicular, and which are neither.

Check the conditions. The problem states that the lines are different and nonvertical, so both slope criteria apply.
Look for equal slopes. Lines P and Q both have slope 3/4, and no other pair matches. By the Parallel Lines Slope Criterion, in the slopes-to-lines direction, equal slopes mean Lines P and Q are parallel.
$m_P = m_Q = \frac{3}{4}$
Multiply slopes to test for perpendicular lines, starting with Lines P and R. The product is -1, so by the Perpendicular Lines Slope Criterion, in the slopes-to-lines direction, they are perpendicular. Line Q has the same slope as Line P, so Lines Q and R are perpendicular too.
$\frac{3}{4} \cdot \left(-\frac{4}{3}\right) = -\frac{3 \cdot 4}{4 \cdot 3} = -\frac{12}{12} = -1$
Test Line S with Line P. The slopes differ and the product is 1, not -1, so the lines-to-slopes direction of each criterion rules out that relationship. This pair is neither. Line Q has the same slope as Line P, so Lines S and Q are neither as well.
$\frac{3}{4} \cdot \frac{4}{3} = \frac{3 \cdot 4}{4 \cdot 3} = \frac{12}{12} = 1$
Test Line S with Line R. The slopes differ and the product is not -1, so by the same reasoning this pair is neither.
$\left(-\frac{4}{3}\right) \cdot \frac{4}{3} = -\frac{4 \cdot 4}{3 \cdot 3} = -\frac{16}{9}$

Answer

Lines P and Q are parallel, and Line R is perpendicular to both of them. Line S is neither parallel nor perpendicular to any of the other three lines.

Mistake: using only half of the perpendicular test. Some learners call lines with slopes $2$ and $-2$ perpendicular because the signs are opposite. Others pick slopes $2$ and $\frac{1}{2}$ because one is the other flipped. Multiply, and both fail: $2 \cdot (-2) = -4$ and $2 \cdot \frac{1}{2} = 1$ . Neither product is $-1$ . Here is why. Turning a line of slope $\frac{2}{1}$ a quarter turn swaps the $2$ and the $1$ and also changes one sign, which gives slope $-\frac{1}{2}$ . Slope $-2$ changes the sign without the swap, and slope $\frac{1}{2}$ swaps without changing the sign, so neither line makes a right angle with a line of slope $2$ . For nonvertical lines, always multiply the slopes and check for exactly $-1$ .

How to decide how two lines are related

What you find	Conclusion	Reason
Both nonvertical and different, with $m_1 = m_2$	Parallel	Parallel Lines Slope Criterion, slopes to lines
Both nonvertical, with $m_1 m_2 = -1$	Perpendicular	Perpendicular Lines Slope Criterion, slopes to lines
Both nonvertical, with $m_1 \neq m_2$ and $m_1 m_2 \neq -1$	Neither	Both criteria, lines to slopes
One vertical and one horizontal	Perpendicular	The coordinate axes are perpendicular, and no product is used
One vertical, the other neither vertical nor horizontal	Neither	A line parallel to a vertical line is vertical; one perpendicular to it is horizontal
Two different vertical lines	Parallel	Their $x$ -coordinates differ, so they share no point (definition of parallel lines)

Worked example

Deciding from four points

Given: Line J passes through $(-1, -2)$ and $(5, 2)$ , and Line K passes through $(4, -3)$ and $(-2, 6)$ . Decide whether the lines are parallel, perpendicular, or neither.

Use the definition of slope for Line J. The change in x is 6, not 0, so Line J is nonvertical. Divide the top and bottom by 2 to simplify.
$m_J = \frac{2 - (-2)}{5 - (-1)} = \frac{2 + 2}{5 + 1} = \frac{4}{6} = \frac{2}{3}$
Do the same for Line K. The change in x is -6, not 0, so Line K is nonvertical. Divide the top and bottom by 3 to simplify.
$m_K = \frac{6 - (-3)}{-2 - 4} = \frac{6 + 3}{-6} = \frac{9}{-6} = -\frac{3}{2}$
Compare the slopes. They are not equal, so the Parallel Lines Slope Criterion, in the lines-to-slopes direction, rules out parallel.
$\frac{2}{3} \neq -\frac{3}{2}$
Multiply the slopes. Both lines are nonvertical, so the Perpendicular Lines Slope Criterion applies.
$\frac{2}{3} \cdot \left(-\frac{3}{2}\right) = -\frac{2 \cdot 3}{3 \cdot 2} = -\frac{6}{6} = -1$
The product is -1, so the slopes-to-lines direction of that criterion tells you the lines are perpendicular.

Answer

Line J and Line K are perpendicular.

Worked example

A vertical line in the mix

Given: Line T passes through $(3, -1)$ and $(3, 4)$ , Line U passes through $(-2, 1)$ and $(2, 3)$ , and Line V passes through $(-2, 1)$ and $(4, 1)$ . Decide whether each pair of lines is parallel, perpendicular, or neither.

Check the change in x first. For Line T it is 0, so Line T is vertical and its slope is undefined.
$x_2 - x_1 = 3 - 3 = 0$
Line U has a nonzero change in x, so use the definition of slope.
$m_U = \frac{3 - 1}{2 - (-2)} = \frac{2}{2 + 2} = \frac{2}{4} = \frac{1}{2}$
Line V also has a nonzero change in x. Its slope is 0, so it is horizontal.
$m_V = \frac{1 - 1}{4 - (-2)} = \frac{0}{4 + 2} = \frac{0}{6} = 0$
Lines T and V: one is vertical and one is horizontal, so they are perpendicular. No product is used, because Line T has undefined slope.
Lines T and U: Line U is nonvertical, so it is not parallel to Line T, since a line parallel to a vertical line is vertical. Its slope is not 0, so it is not horizontal, and only a horizontal line is perpendicular to a vertical line. This pair is neither.
$m_U = \frac{1}{2} \neq 0$
Lines U and V: both are nonvertical, so the criteria apply. The slopes differ and the product is not -1, so the lines-to-slopes direction of each criterion rules out that relationship. This pair is neither.
$\frac{1}{2} \neq 0, \qquad \frac{1}{2} \cdot 0 = 0 \neq -1$

Answer

Lines T and V are perpendicular, Lines T and U are neither parallel nor perpendicular, and Lines U and V are neither parallel nor perpendicular.

Takeaway: slope is the change in $y$ divided by the change in $x$ , for a nonzero change in $x$ , and it lets you compare lines with numbers instead of judging by how a picture looks. Two different nonvertical lines are parallel exactly when their slopes are equal and perpendicular exactly when the product of their slopes is $-1$ . A vertical line and a horizontal line are perpendicular without a product. Next, you will graph a line through a point parallel or perpendicular to a given line, and you will prove these slope criteria.