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In this lesson, the relationship between the slopes of parallel and perpendicular lines will be explored.

Catch-Up and Review

Here are a few recommended readings before getting started with this lesson.

Lines
Slope of a Line
Linear Equations
System of Linear Equations
Solving a System of Linear Equations
Similarity
Right Triangles
Trigonometry

Explore

Investigating the Slopes of Parallel Lines

The lines $ℓ_{1}$ and $ℓ_{2}$ are positioned as shown in the graph. Move the point $C$ vertically.

Compare the slope triangles. What conclusion can be made about the triangles and the lines?

Discussion

Slopes of Parallel Lines Theorem

In the previous graph, the slope triangles $△ A B F$ and $△ C D E$ can be mapped onto each other by a translation. Since translations are rigid motions, it can be concluded that $△ A B F$ and $△ C D E$ are congruent triangles. Because the slope triangles are congruent, the slopes of the lines are equal. This means that the lines are parallel.

In a coordinate plane, two distinct non-vertical lines are parallel if and only if their slopes are equal.

If $ℓ_{1}$ and $ℓ_{2}$ are two parallel lines and $m_{1}$ and $m_{2}$ their respective slopes, then the following statement is true.

$ℓ_{1} ∥ ℓ_{2} \Leftrightarrow m_{1} = m_{2}$

The slope of a vertical line is not defined. Therefore, this theorem only applies to non-vertical lines. However, any two distinct vertical lines are parallel.

Proof

Since the theorem consists of a biconditional statement, the proof consists of two parts.

If two distinct non-vertical lines are parallel, then their slopes are equal.
If the slopes of two distinct non-vertical lines are equal, then the lines are parallel.

Part $1$

Consider two distinct non-vertical parallel lines in a coordinate plane. Their equations can be written in slope-intercept form.

ℓ_{1} : y = m_{1} x + b_{1} ℓ_{2} : y = m_{2} x + b_{2}

Suppose that the slopes of the lines are not the same. The system of equations formed by the equations above can be solved by using the Substitution Method.

{y = m_{1} x + b_{1} y = m_{2} x + b_{2} (I) (II)

Substitute

$(I):$ $y = m_{2} x + b_{2}$

{m_{2} x + b_{2} = m_{1} x + b_{1} y = m_{2} x + b_{2}

▼

(I):

Solve for

x

SubEqn

$(I):$ $LHS - b_{2} = RHS - b_{2}$

{m_{2} x = m_{1} x + b_{1} - b_{2} y = m_{2} x + b_{2}

SubEqn

$(I):$ $LHS - m_{1} x = RHS - m_{1} x$

{m_{2} x - m_{1} x = b_{1} - b_{2} y = m_{2} x + b_{2}

FactorOut

$(I):$ Factor out $x$

{(m_{2} - m_{1}) x = b_{1} - b_{2} y = m_{2} x + b_{2}

DivEqn

$(II):$ $LHS / (m_{2} - m_{1}) = RHS / (m_{2} - m_{1})$

{x = \frac{b _{1} - b _{2}}{m _{2} - m _{1}} y = m_{2} x + b_{2}

Since

m_{1} \neq = m_{2},

the expression

\frac{b _{1} - b _{2}}{m _{2} - m _{1}}

is not undefined because its denominator cannot be zero. To find the value of the

y -

variable,

\frac{b _{1} - b _{2}}{m _{2} - m _{1}}

can be substituted for

x

in Equation (II).

{x = \frac{b _{1} - b _{2}}{m _{2} - m _{1}} y = m_{2} x + b_{2}

Substitute

$(II):$ $x = \frac{b _{1} - b _{2}}{m _{2} - m _{1}}$

{x = \frac{b _{1} - b _{2}}{m _{2} - m _{1}} y = m_{2} (\frac{b _{1} - b _{2}}{m _{2} - m _{1}}) + b_{2}

The solution to the system formed by the equations was found. Since there is a solution for the system, the lines

ℓ_{1}

and

ℓ_{2}

intersect each other. However, this contradicts the fact that the lines are parallel. Therefore, the assumption that the slopes are different is false. Consequently, the slopes of the lines are equal.

$ℓ_{1} ∥ ℓ_{2} \Rightarrow m_{1} = m_{2}$

Part $2$

Now, consider two distinct non-vertical lines

ℓ_{1}

and

ℓ_{2}

that have the same slope

m .

Their equations can be written in slope-intercept form.

ℓ_{1} : y = m x + b_{1} ℓ_{2} : y = m x + b_{2}

Since these lies are distinct,

b_{1}

and

b_{2}

are not equal. With this information in mind, suppose that the lines intersect. Solving the system of equations will give the point of intersection. The Substitution Method will be used again.

{y = m x + b_{1} y = m x + b_{2} (I) (II)

Substitute

$(I):$ $y = m x + b_{2}$

{m x + b_{2} = m x + b_{1} y = m x + b_{2}

SubEqn

$(I):$ $LHS - m x = RHS - m x$

{b_{2} = b_{1} y = m x + b_{2}

The obtained result contradicts the fact that

b_{1}

and

b_{2}

are different. Therefore, there is no point of intersection between the lines

ℓ_{1}

and

ℓ_{2} .

This means that they are parallel lines.

$m_{1} = m_{2} \Rightarrow ℓ_{1} ∥ ℓ_{2}$

Both directions of the biconditional statement have been proved.

$ℓ_{1} ∥ ℓ_{2} \Leftrightarrow m_{1} = m_{2}$

Pop Quiz

Practice Finding the Slope of Parallel lines

If the equation of a linear function is written in slope-intercept form, its slope can be identified.

By rewriting the given equation in slope-intercept form, find the slope of a parallel line to the line whose equation is shown. If necessary, round your answer to $2$ decimal places.

Example

Identifying the Equation of a Parallel Line

Consider the equation of a line written in slope-intercept form.

y = - 2 x - 5

Which of the following is the equation of the line that passes through the point

(2, - 3)

and is parallel to the given line.

Hint

Start by identifying the slope of the given line. Then, use the Slopes of Parallel Lines Theorem.

Solution

Consider both the general form of the slope-intercept form of a line and the given line.

Slope - Intercept Form : Given Line : y = m x + b y = - 2 x - 5

In the slope-intercept form,

m

represents the slope and

b

the

y -

intercept. Since the given line is in this form, the slope of the line is

- 2 .

By the Slopes of Parallel Lines Theorem, a parallel line to this line also has a slope of

- 2 .

Given Line : Parallel Line : y = - 2 x - 5 y = - 2 x + b

It is given that this parallel line passes through the point

(2, - 3) .

By substituting this point into the equation of the parallel line, its

y -

intercept can be found.

y = - 2 x + b

SubstituteII

$x = 2$ , $y = - 3$

- 3 = - 2 (2) + b

▼

Solve for

b

MultNegPos

$(- a) b = - a b$

- 3 = - 4 + b

AddEqn

$LHS + 4 = RHS + 4$

1 = b

RearrangeEqn

Rearrange equation

b = 1

The equation of the parallel line to

y = - 2 x - 5

through the point

(2, - 3)

y = - 2 x + 1 .

Given Line : Parallel Line : y = - 2 x - 5 y = - 2 x + 1

Example

Finding a Parallel Line When Given a Graph and a Point

Up to now, it has been discussed how to find the equation of a parallel line to a line whose equation is given. What about finding the equation of a parallel line to a line whose graph is given?

The first transatlantic telegraph cable was laid between Valentia in western Ireland and Trinity Bay Newfoundland in the $1850 s .$ With the invention of fiber optic cables, the number of transatlantic cables has increased. The following map shows some of these cables.

Kevin wants to write the equation of a line parallel to the first transatlantic cable and passes through point

V (- 4, - 3),

a specific location in Virginia beach. Kevin draws a coordinate plane, then line

ℓ_{1},

on which the cable lies, and the point

V

as shown.

a If Kevin wants to use transformations to find the equation of the parallel line, what would be his next step? Use transformations to write the equation of the parallel line. Write the equation in slope-intercept form.

b Use the Slopes of Parallel Lines Theorem to find the equation of the line. Write the equation in slope-intercept form.

Answer

a Example Next Step: The next step would be determining the number of units needed to vertically translate the given line so that it passes through the given point.

y = \frac{1}{4} x - 2

b Equation:

y = \frac{1}{4} x - 2

Hint

a By which transformation can a line be mapped onto a parallel line?

b What is the slope of the given line?

Solution

a Parallel lines can be mapped onto each other by a translation. Therefore, the next step will be to determine the number of units needed to translate the line so that it passes through

V .

As can be seen in the graph, if the given line is translated

5

units down, a parallel line through

V (- 4, - 3)

is obtained. As a result, the

y -

values of the parallel line will be

5

less than the corresponding

y -

values of the given line. With this knowledge, the equation of the parallel line can be found by following two steps.

Find the equation of the given line.
Translate the given line $5$ units down.

Step $1$

By observing the graph, it can be seen that the line $ℓ_{1}$ passes through the points $(0, 3)$ and $(4, 4) .$ When a slope triangle is constructed between these points, the slope of the line is calculated as $\frac{1}{4} .$

The graph of a straight line through the points (0,3) and (4,4) with a slope of 1/4.

It can be seen that the line intercepts the

y -

axis at

(0, 3) .

Therefore, the

y -

intercept is

3 .

Knowing the slope and the

y -

intercept of line is enough to write its equation in slope-intercept form.

ℓ_{1} : y = \frac{1}{4} x + 3

Step $2$

Recall that the

y -

values of the parallel line are

5

less than the corresponding

y -

values of the given line. Therefore, to obtain the equation of the desired line,

5

units must be subtracted from the obtained equation.

\underline{Given Line} y = f (x) ⇓ y = \frac{1}{4} x + 3 \underline{Parallel Line} y = f (x) - 5 ⇓ y = \frac{1}{4} x + 3 - 5 y = \frac{1}{4} x - 2

The equation of the parallel line to

ℓ_{1}

through

V (- 4, - 3)

y = \frac{1}{4} x - 2 .

b Using the points

(0, 3)

and

(4, 4),

the slope of the line is calculated as

\frac{1}{4} .

Therefore, by the Slopes of Parallel Lines Theorem, all parallel lines to

ℓ_{1}

have a slope of

\frac{1}{4} .

Then, the equation of the parallel line passing through

P

can be written as follows, where

b

is the

y -

intercept.

y = \frac{1}{4} x + b

Since

V (- 4, - 3)

should lie on the line, the value of

b

can be found by substituting its coordinates into the above equation.

y = \frac{1}{4} x + b

SubstituteII

$x = - 4$ , $y = - 3$

- 3 = \frac{1}{4} (- 4) + b

▼

Solve for

b

MoveRightFacToNumOne

$\frac{1}{b} \cdot a = \frac{a}{b}$

- 3 = \frac{- 4}{4} + b

MoveNegNumToFrac

Put minus sign in front of fraction

- 3 = - \frac{4}{4} + b

QuotOne

$\frac{a}{a} = 1$

- 3 = - 1 + b

AddEqn

$LHS + 1 = RHS + 1$

- 2 = b

RearrangeEqn

Rearrange equation

b = - 2

The equation of the parallel line to

ℓ_{1}

that passes through

V (- 4, - 3)

y = \frac{1}{4} x - 2 .

The equation is the same as the equation obtained in Part A, as expected.

Example

Systems of Equations With Parallel or Overlapping Lines

Recall that a system of linear equations can have infinitely many solutions, one solution or no solution. Under which conditions does a system of linear equations have infinitely many solutions or no solution?

Consider the following systems of equations.

System I : System II : {y = 3 x - 3 2 y = 6 x - 6 (I) (II) {y = - 5 x - 3 y = - 5 x - 1 (I) (II)

Determine the number of solutions to each system without finding the actual solutions, if exist.

Answer

System I: Infinitely many solutions
System II: No solution

Hint

Determine whether the equations in the system represent parallel lines. What does this say about the number of solutions?

Solution

Start by examining the first system of equations.

System I : {y = 3 x - 3 2 y = 6 x - 6 (I) (II)

Dividing both sides of the second equation by

2

will result in the first equation. This result means that the equations in this system represent the same line.

{y = 3 x - 3 2 y = 6 x - 6 (II) \div 2 {y = 3 x - 3 y = 3 x - 3

Therefore, the lines are coincidental, and they have infinitely many points of intersection. Consequently, System (I) has infinitely many solutions. The equations in the System (II), however, differ in their

y -

intercepts.

System II : {y = - 5 x - 3 y = - 5 x - 1 (I) (II)

Since the equations are written in slope-intercept form, their slopes can be identified as

- 5 .

Equation	Slope	$y -$ Intercept
$y = - 5 x + (- 3)$	$- 5$	$- 3$
$y = - 5 x + (- 1)$	$- 5$	$- 1$

By the Slopes of Parallel Lines Theorem, these lines are parallel and do not intersect. Therefore, System (II) has no solution.

Discussion

Properties of Different Systems of Equations

From the previous example, the following conclusions about systems of linear equations can be made.

Condition	Conclusion	Example
The lines of the system have the same slope and the same $y -$ intercept.	The lines are coincidental. This means that there are infinitely many points of intersection. Therefore, the system has infinitely many solutions.	${y = 3 x + 1 y = 3 x + 1$
The lines of the system have the same slope but different $y -$ intercept.	The lines are parallel. This means that there is not a point of intersection. Therefore, the system has no solution.	${y = 3 x + 1 y = 3 x + 5$
The lines of the system have different slopes.	The lines are neither parallel nor coincidental. This means that there is one point of intersection. Therefore, the system has one solution.	${y = 3 x + 1 y = - 2 x + 5$

These conclusions can be seen in the following diagram.

Explore

Investigating Lines Using Slope Triangles

In the graph below, the slope triangles seem to be congruent. The congruence of these triangles can be shown by a rotation. Move the slider to rotate line

ℓ_{1}

counterclockwise around point

A .

Think about these questions.

Is it possible to find a relationship between the slopes of the lines before rotating $ℓ_{1} ?$
How does the rotation change the slope of line $ℓ_{1} ?$

Discussion

Properties of Perpendicular Lines

In the previous graph, it can be seen that the initial angle between the lines measures $9 0^{\circ} .$ Therefore, the lines are perpendicular.

Rule

Slopes of Perpendicular Lines Theorem

In a coordinate plane, two non-vertical lines are perpendicular if and only if their slopes are negative reciprocals.

If $ℓ_{1}$ and $ℓ_{2}$ are two perpendicular lines and $m_{1}$ and $m_{2}$ their respective slopes, the following relation holds true.

$ℓ_{1} ⊥ ℓ_{2} \Leftrightarrow m_{1} \cdot m_{2} = - 1$

This theorem does not apply to vertical lines because their slope is undefined. However, vertical lines are always perpendicular to horizontal lines.

Proof

Since the theorem is a biconditional statement, the proof consists of two parts.

If two non-vertical lines are perpendicular, then the product of their slopes is $- 1 .$
If the product of the slopes of two non-vertical lines is $- 1,$ then the lines are perpendicular.

Part $1$

Let

ℓ_{1}

and

ℓ_{2}

be two perpendicular lines. Therefore, they intersect at one point. For simplicity, the lines will be translated so that the point of intersection is the origin.

Let

m_{1}

and

m_{2}

be the slopes of the lines

ℓ_{1}

and

ℓ_{2},

respectively. Next, consider the vertical line

x = 1 .

This line intersects both

ℓ_{1}

and

ℓ_{2} .

Since

ℓ_{1}

and

ℓ_{2}

are assumed to be perpendicular,

△ A O C

is a right triangle. Using the Distance Formula, the lengths of the sides of this triangle can be found.

Side	Points	$Distance Formula (x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}$	Length
$\overline{A O}$	$A (1, m_{1})$ $&$ $O (0, 0)$	$(0 - 1)^{2} + (0 - m_{1})^{2}$	$1 + m_{1}^{2}$
$\overline{C O}$	$C (1, m_{2})$ $&$ $O (0, 0)$	$(0 - 0)^{2} + (0 - m_{2})^{2}$	$1 + m_{2}^{2}$
$\overline{C A}$	$C (1, m_{2})$ $&$ $A (1, m_{1})$	$(1 - 1)^{2} + (m_{1} - m_{2})^{2}$	$m_{1} - m_{2}$

Since

△ A O C

is a right triangle, its side lengths satisfy the Pythagorean Equation.

A O^{2} + C O^{2} = C A^{2}

The next step is to substitute the lengths shown in the table.

A O^{2} + C O^{2} = C A^{2}

SubstituteExpressions

Substitute expressions

(1 + m_{1}^{2})^{2} + (1 + m_{2}^{2})^{2} = (m_{1} - m_{2})^{2}

▼

Simplify

PowSqrt

$(a)^{2} = a$

1 + m_{1}^{2} + 1 + m_{2}^{2} = (m_{1} - m_{2})^{2}

AddTerms

Add terms

2 + m_{1}^{2} + m_{2}^{2} = (m_{1} - m_{2})^{2}

ExpandNegPerfectSquare

$(a - b)^{2} = a^{2} - 2 a b + b^{2}$

2 + m_{1}^{2} + m_{2}^{2} = m_{1}^{2} - 2 m_{1} m_{2} + m_{2}^{2}

SubEqn

$LHS - m_{1}^{2} = RHS - m_{1}^{2}$

2 + m_{2}^{2} = - 2 m_{1} m_{2} + m_{2}^{2}

SubEqn

$LHS - m_{2}^{2} = RHS - m_{2}^{2}$

2 = - 2 m_{1} m_{2}

DivEqn

$LHS / (- 2) = RHS / (- 2)$

\frac{2}{- 2} = m_{1} m_{2}

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Catch-Up and Review

Proof

Part 1

Part 2

Hint

Solution

Answer

Hint

Solution

Step 1

Step 2

Answer

Hint

Solution

Proof

Part 1

Part $1$

Part $2$

Step $1$

Step $2$

Part $1$