Graph and Prove Parallel and Perpendicular Slopes | Geometry | LevelKite

You already use slopes to decide whether two lines are parallel, perpendicular or neither. In this lesson you use slopes to graph such a line through a chosen point, and then you prove why the slope tests work.

Three definitions carry the lesson. Slope is the change in $y$ over the change in $x$ , for a nonzero change in $x$ ; a vertical line has undefined slope, and a horizontal line has slope $0$ . Parallel lines are lines in the same plane that never meet. Perpendicular lines are lines that meet to form a right angle; one right angle at the meeting point makes all four angles there right angles.

Think of a slope as walking directions along a line. The run is the change in $x$ and the rise is the change in $y$ , so a slope of $\frac{3}{2}$ says right 2, up 3.

Parallel lines tilt the same way. When a horizontal line crosses two parallel lines that are not horizontal, the corresponding angles are congruent, so both lines follow the same directions. A perpendicular line is a quarter turn away. Turn right 2, up 3 a quarter turn to the left and you get left 3, up 2, a slope of $\frac{2}{-3} = -\frac{2}{3}$ . Notice that $\frac{3}{2} \cdot \left(-\frac{2}{3}\right) = -1$ . This picture gives you the idea; the proofs later in this lesson make it certain.

Key concept

The slope criteria

Take two different nonvertical lines in the coordinate plane, with slopes $m_1$ and $m_2$ .

Slope Criterion for Parallel Lines: the lines are parallel exactly when $m_1 = m_2$ .
Slope Criterion for Perpendicular Lines: the lines are perpendicular exactly when $m_1m_2 = -1$ .

Exactly when means the statement works in both directions. A vertical line has undefined slope, so these tests do not apply to it. A vertical line and a horizontal line are perpendicular without a product.

m_1 = m_2 \ \text{(parallel)} \qquad m_1 m_2 = -1 \ \text{(perpendicular)}

To graph a line through a given point when the given line is not vertical:

Find the slope of the given line.
Choose the new slope. A parallel line has the same slope, and the point must not be on the given line, or you would only redraw that line. A perpendicular line has the slope whose product with the given slope is $-1$ , so the given slope cannot be $0$ .
Plot the point, then use the run and rise of the new slope to plot a second point.
Draw the line through both points. Check its slope from the coordinates, never from how the graph looks.

In each figure in this lesson, the horizontal axis is the $x$ -axis and the vertical axis is the $y$ -axis. In the two graphing figures, Line A is the given line and Line B is the line you graph.

Worked example

A parallel line

Graph the line through $(1, 4)$ that is parallel to $y = 2x - 3$ .

Read the slope of the given line. In the form y = mx + b, the slope is the number multiplying x, so the slope here is 2.
$y = 2x - 3 \;\Rightarrow\; m_1 = 2$
Choose the new slope. If a line is parallel to a nonvertical line, then it has the same slope, so the new slope is also 2.
$m_2 = m_1 = 2$
Check that the point is not on the given line, so that you get a second line. Substituting x = 1 into the given equation gives y = -1, not 4.
$2(1) - 3 = -1 \neq 4$
Plot (1, 4). Write the slope 2 as 2 over 1, a run of 1 and a rise of 2. Move right 1 and up 2 to reach a second point.
$(1 + 1,\ 4 + 2) = (2, 6)$
Draw the line through the two points and check its slope from the coordinates. Here you use the other direction of the criterion: two different nonvertical lines with equal slopes are parallel.
$\frac{6 - 4}{2 - 1} = \frac{2}{1} = 2$

Answer

The graph is the line through $(1, 4)$ and $(2, 6)$ . It is parallel to $y = 2x - 3$ because the two lines are different and both have slope $2$ .

Line A is

y = 2x - 3

. Line B passes through

(1, 4)

and

(2, 6)

. The triangle marks the run of

1

and the rise of

2

.

Worked example

A perpendicular line

Graph the line through $(3, 4)$ that is perpendicular to $3x - 2y = -2$ .

Solve the given equation for y so you can read its slope. Subtract 3x from both sides, then divide both sides by -2. The given slope is 3/2.
$3x - 2y = -2 \;\Rightarrow\; -2y = -3x - 2 \;\Rightarrow\; y = \frac{3}{2}x + 1$
The given line is nonvertical and its slope is not 0, so look for the slope whose product with 3/2 is -1. Multiply both sides by 2/3.
$\frac{3}{2}m_2 = -1 \;\Rightarrow\; m_2 = -1 \cdot \frac{2}{3} = -\frac{2}{3}$
Plot (3, 4). Read the new slope as a run of 3 and a rise of -2, so move right 3 and down 2.
$(3 + 3,\ 4 - 2) = (6, 2)$
Draw the line through the two points. Check its slope from the coordinates, then check the product. This uses the direction from the product to the lines: if two nonvertical lines have slopes with product -1, then they are perpendicular.
$\frac{2 - 4}{6 - 3} = -\frac{2}{3}, \qquad \frac{3}{2} \cdot \left(-\frac{2}{3}\right) = -1$

Answer

The graph is the line through $(3, 4)$ and $(6, 2)$ . It is perpendicular to $3x - 2y = -2$ because both lines are nonvertical and the product of their slopes is $-1$ .

Line A is

3x - 2y = -2

, with slope

\frac{3}{2}

. Line B passes through

(3, 4)

and

(6, 2)

. The triangle marks the run of

3

and the rise of

-2

.

The product test cannot help in two cases: a vertical given line has undefined slope, and a horizontal given line has slope $0$ , and no number times $0$ equals $-1$ . In these cases you use the line's direction instead.

Take the vertical line $x = 4$ and the point $(1, 3)$ . The line through the point parallel to $x = 4$ is the vertical line $x = 1$ . No point has two different $x$ -coordinates, so these lines share no point, and they are parallel by the Definition of parallel lines. The line through the point perpendicular to $x = 4$ is the horizontal line $y = 3$ , because a vertical line and a horizontal line are perpendicular. You will prove that fact at the end of this lesson. The table collects every case for a point $(h, k)$ .

Graphing a line through the point (h, k)

Given line	Parallel line	Perpendicular line
Nonvertical, slope $m \neq 0$	Slope $m$ , if $(h, k)$ is not on the given line	Slope $-\frac{1}{m}$
Horizontal, $y = c$ , slope $0$	$y = k$ , if $k \neq c$	Vertical line $x = h$
Vertical, $x = c$ , undefined slope	$x = h$ , if $h \neq c$	Horizontal line $y = k$

Now you prove the criteria. Exactly when packs in two statements: a conditional (if this, then that) and its converse, which swaps the if part and the then part. A statement can be true while its converse is false, so you must prove both directions.

In a proof, the Given is what you start from, and the statement to prove is what you must reach. Each reason below is a definition, a property, a proved theorem or the word Given. Nothing is read from a graph.

You need one more definition. The graph of an equation is the set of all points whose coordinates make the equation true. Every nonvertical line is the graph of an equation $y = mx + b$ , where $m$ is its slope and $b$ is its $y$ -intercept.

Worked example

Proof of the Slope Criterion for Parallel Lines

Given: two different nonvertical lines in the coordinate plane, $y = m_1x + b_1$ and $y = m_2x + b_2$ . Prove: the lines are parallel exactly when $m_1 = m_2$ .

Find what a shared point must satisfy. A point on both lines makes both equations true (Definition of the graph of an equation). Both equations then give the same y, so their right sides are equal (Substitution Property of Equality).
$m_1x + b_1 = m_2x + b_2$
Direction 1: if the slopes are equal, then the lines are parallel. The equal slopes are Given here. Replace the second slope with the first (Substitution Property of Equality), then subtract the same term from both sides (Subtraction Property of Equality). A shared point would need equal intercepts.
$m_1x + b_1 = m_1x + b_2 \;\Rightarrow\; b_1 = b_2$
Equal slopes and equal intercepts would give one equation and so one graph (Definition of the graph of an equation), but the lines are different (Given). So the intercepts are not equal, no point lies on both lines, and the lines are parallel (Definition of parallel lines).
$b_1 \neq b_2$
Direction 2, the converse: if the lines are parallel, then the slopes are equal. The parallel lines are Given here, so they share no point (Definition of parallel lines). Now see what different slopes would do. Their difference would not be 0, so this fraction would have a nonzero divisor, and x would be a real number.
$x = \frac{b_2 - b_1}{m_1 - m_2}, \quad m_1 - m_2 \neq 0$
Multiply both sides by the difference of the slopes (Multiplication Property of Equality) and distribute (Distributive Property). Then add the second line's x-term and the first intercept to both sides (Addition Property of Equality). This x gives one y on both lines, so that point is on both graphs (Definition of the graph of an equation).
$(m_1 - m_2)x = b_2 - b_1 \;\Rightarrow\; m_1x - m_2x = b_2 - b_1 \;\Rightarrow\; m_1x + b_1 = m_2x + b_2$
So different slopes would produce a shared point, but parallel lines have none (Definition of parallel lines). The slopes cannot be different, so they are equal.
$m_1 = m_2$

Answer

Two different nonvertical lines are parallel exactly when their slopes are equal. This is the Slope Criterion for Parallel Lines.

The perpendicular proof uses three tools you already know. The Distance Formula says the square of the distance between $(x_1, y_1)$ and $(x_2, y_2)$ is $(x_2 - x_1)^2 + (y_2 - y_1)^2$ . The Pythagorean Theorem says that if a triangle is a right triangle with legs $a$ and $b$ and hypotenuse $c$ (the side opposite the right angle), then $c^2 = a^2 + b^2$ . The Converse of the Pythagorean Theorem says that if the sides of a triangle satisfy $c^2 = a^2 + b^2$ , then the angle opposite side $c$ is a right angle.

The figure tracks the letters in one case, with slopes $2$ and $-\frac{1}{2}$ . Point C is $(1, 1)$ . One unit to its right on each line are Point A, $(2, 3)$ , and Point B, $(2, \frac{1}{2})$ . These coordinates give $AC^2 = 5$ , $BC^2 = \frac{5}{4}$ and $AB^2 = \frac{25}{4}$ , and $\frac{25}{4} = 5 + \frac{5}{4}$ . One case is not a proof, so the proof below uses letters for the slopes.

Line A has slope

2

and Line B has slope

-\frac{1}{2}

. They meet at Point C,

(1, 1)

. The triangle is triangle

ABC

from the proof.

Worked example

Proof of the Slope Criterion for Perpendicular Lines

Given: two nonvertical lines in the coordinate plane, Line A with slope $m_1$ and Line B with slope $m_2$ . Prove: the lines are perpendicular exactly when $m_1m_2 = -1$ .

Set up the points for lines that meet at a point C. By the Definition of slope, a run of 1 along a line gives a rise equal to its slope. So moving 1 unit right of C along each line gives Point A on Line A and Point B on Line B.
$C(p, q), \quad A(p + 1,\ q + m_1), \quad B(p + 1,\ q + m_2)$
Find the squared lengths (Distance Formula). From C to A or to B, the change in x is 1 and the change in y is that line's slope. From B to A, the change in x is 0 and the change in y is the difference of the slopes.
$AC^2 = 1^2 + m_1^2, \quad BC^2 = 1^2 + m_2^2, \quad AB^2 = 0^2 + (m_1 - m_2)^2$
Link the Pythagorean equation to the slopes. Substitute the squared lengths (Substitution Property of Equality) and expand (Distributive Property). Subtract the squared slopes (Subtraction Property of Equality), then divide by -2 (Division Property of Equality). Each double arrow works both ways; going up uses the Multiplication and Addition Properties of Equality.
$\begin{aligned} & AB^2 = AC^2 + BC^2 \\ \iff\ & m_1^2 - 2m_1m_2 + m_2^2 = 2 + m_1^2 + m_2^2 \\ \iff\ & -2m_1m_2 = 2 \\ \iff\ & m_1m_2 = -1 \end{aligned}$
Direction 1: if the lines are perpendicular, then the product is -1. The perpendicular lines are Given here. They meet at C and angle ACB is a right angle (Definition of perpendicular lines), so AB is the hypotenuse of right triangle ABC. The Pythagorean Theorem gives the top line of the chain, and the arrows lead down to the product.
$\angle ACB = 90^\circ \;\Rightarrow\; AB^2 = AC^2 + BC^2 \;\Rightarrow\; m_1m_2 = -1$
Direction 2, the converse: if the product is -1, then the lines are perpendicular. The product of -1 is Given here. Equal slopes would make the product a square, and the square of a real number is never negative. So the slopes differ, the lines are not parallel (Slope Criterion for Parallel Lines), and they meet (Definition of parallel lines). Name C, A and B as in the first step.
$m_1 = m_2 \;\Rightarrow\; m_1m_2 = m_1^2 \geq 0, \text{ so } m_1m_2 \neq -1$
A and B are different points on a vertical line, because the slopes differ, and C is not on that line, so ABC is a triangle. Reading the chain upward gives the top equation. By the Converse of the Pythagorean Theorem, angle ACB, opposite AB, is a right angle. So the lines are perpendicular (Definition of perpendicular lines).
$m_1m_2 = -1 \;\Rightarrow\; AB^2 = AC^2 + BC^2 \;\Rightarrow\; \angle ACB = 90^\circ$

Answer

Two nonvertical lines are perpendicular exactly when the product of their slopes is $-1$ . This is the Slope Criterion for Perpendicular Lines.

The vertical and horizontal case needs its own proof, because a vertical line has undefined slope and there is no product to take. The same triangle idea works. Let the horizontal line $y = k$ and the vertical line $x = h$ meet at $C(h, k)$ . Take $A(h + 1, k)$ on the horizontal line and $B(h, k + 1)$ on the vertical line. By the Distance Formula, $AC^2 = 1$ , $BC^2 = 1$ and $AB^2 = 1^2 + (-1)^2 = 2$ . Because $2 = 1 + 1$ , the Converse of the Pythagorean Theorem makes angle $ACB$ a right angle. So the lines are perpendicular by the Definition of perpendicular lines.

The takeaway: slope tells you the direction of a line, so for a nonvertical line, a point and a slope are all you need to graph it. For two different nonvertical lines, equal slopes and parallel lines always go together, and a slope product of $-1$ and perpendicular lines always go together. You proved both directions of each criterion from definitions, properties and named theorems, not from how a graph looks. A vertical line and a horizontal line are perpendicular without a product.