Understanding Parallel and Perpendicular Lines: Equations Explored

Two different lines on the same plane can either intersect or not. If the lines do not intersect, then they are parallel. If the lines do intersect, they might be perpendicular. It can be useful to know how to identify parallel and perpendicular lines, which will be explored in this lesson.

Catch-Up and Review

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

Slope of a Line
Slope Formula
$y$ -intercept
Slope-Intercept Form
Graphing a Linear Function in Slope-Intercept Form

Explore

Observing Parallel Lines

Consider the following applet showing two parallel lines and their equations in slope-intercept form.

What characteristic of the lines remains unchanged when they are moved up or down?

Discussion

Parallel Lines

Two coplanar lines — lines that are on the same plane — that do not intersect are said to be parallel lines. In a diagram, triangular hatch marks are drawn on lines to denote that they are parallel. The symbol $∥$ is used to algebraically denote that two lines are parallel. In the diagram, lines $m$ and $ℓ$ are parallel.

Rule

Slopes of Parallel Lines Theorem

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

Identifying Parallel Lines

Consider two lines on the same coordinate plane and their equations in slope-intercept form. Are the lines parallel?

Example

Finding a Parallel Line Through a Point

Write an equation in slope-intercept form of the line that passes through $(- 2, 3)$ and is parallel to $y = 2 x - 1 .$

Hint

What do parallel lines have in common?

Solution

Consider the given equation of the given line.

y = 2 x - 1

Parallel lines have the same slope. Therefore, the slope of all lines parallel to the given line is

2 .

A general equation in slope-intercept form for these lines can be written.

y = 2 x + b

The desired parallel line passes through

(- 2, 3) .

By substituting the coordinates of this point into the above equation for

x

and

y,

the

y -

intercept

b

of the parallel line can be determined.

y = 2 x + b

SubstituteII

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

3 = 2 (- 2) + b

▼

Solve for

b

MultPosNeg

$a (- b) = - a \cdot b$

3 = - 4 + b

AddEqn

$LHS + 4 = RHS + 4$

7 = b

RearrangeEqn

Rearrange equation

b = 7

Knowing the

y -

intercept, the equation of the line parallel to

y = 2 x - 1

through

(- 2, 3)

can be written.

y = 2 x + 7

The solution can be verified by graphing the line on the same coordinate plane.

As seen above, the graph of $y = 2 x + 7$ passes through $(- 2, 3)$ and is parallel to the graph of $y = 2 x - 1 .$

Example

Bike Path

Kevin's class is entering a contest to redesign parts of their town. They are using new software to create a model of their ideas for changes to the park. Kevin wants to propose a new bike path in the park. The bike path has to be parallel to an already existing pavement. He visualized the situation on the following graph.

A point can be seen on the graph. Kevin wants the bike path to pass through the point, as his measurements show that this is the most convenient option. What is the equation of the line, written in slope-intercept form, that corresponds to the bike path?

Hint

Find the slope of the line shown on the graph.

Solution

Two parallel lines have the same slope. To find the equation of the line that corresponds to the bike path, the slope of the line that represents the pavement has to be determined first. This can be done by using the Slope Formula.

m = \frac{y _{2} - y _{1}}{x _{2} - x _{1}}

In the above formula,

(x_{1}, y_{1})

and

(x_{2}, y_{2})

are two points on the line. Therefore, two points on the line corresponding to the pavement have to be identified.

Now

(0, 3)

and

(1, 1)

will be substituted for

(x_{1}, y_{1})

and

(x_{2}, y_{2}),

respectively.

m = \frac{y _{2} - y _{1}}{x _{2} - x _{1}}

SubstitutePoints

Substitute $(0, 3)$ & $(1, 1)$

m = \frac{1 - 3}{1 - 0}

SubTerms

Subtract terms

m = \frac{- 2}{1}

DivByOne

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

m = - 2

The slope of the pavement is

- 2 .

As mentioned before, parallel lines have the same slope. Therefore, all lines parallel to the line that corresponds to the pavement will have a slope of

- 2 .

A general equation in slope-intercept form for these lines can be written.

y = - 2 x + b

The line that represents the bike path passes through

(- 4, 3.5) .

By substituting the coordinates of this point into the above equation for

x

and

y,

the

y -

intercept

b

of the parallel line can be determined.

y = - 2 x + b

SubstituteII

$x = - 4$ , $y = 3.5$

3.5 = - 2 (- 4) + b

▼

Solve for

b

MultNegNegOnePar

$- a (- b) = a \cdot b$

3.5 = 8 + b

SubEqn

$LHS - 8 = RHS - 8$

- 4.5 = b

RearrangeEqn

Rearrange equation

b = - 4.5

Knowing the

y -

intercept, the equation of the line parallel to the pavement through

(- 4, 3.5)

can be written.

y = - 2 x + (- 4.5) \Leftrightarrow y = - 2 x - 4.5

The solution will be verified by plotting the line on Kevin's graph.

As seen above, the graph of $y = - 2 x - 4.5$ is parallel to the pavement and passes through $(- 4, 3.5) .$

Example

Starry Scene

Zosia wants to propose a new mural to be painted on the side of the planetarium. She starts with a moon and two stars that are already painted on the building.

Zosia wants to place more stars in the line that connects the two existing stars. She also wants to make a second line of stars that is parallel to the first and passes through the moon. The two stars and the moon can be represented on a coordinate plane.

Write the equation of the line of stars that passes through the moon. Give the answer in slope-intercept form.

Hint

Find the slope of the line that passes through the stars.

Solution

The coordinates of the points that represent the stars are

(- 4, - 3)

and

(3, 0.5) .

Since two points on the line are known, the line's slope can be found using the Slope Formula.

m = \frac{y _{2} - y _{1}}{x _{2} - x _{1}}

SubstitutePoints

Substitute $(- 4, - 3)$ & $(3, 0.5)$

m = \frac{0 . 5 - ( - 3 )}{3 - ( - 4 )}

▼

Simplify right-hand side

SubNeg

$a - (- b) = a + b$

m = \frac{0 . 5 + 3}{3 + 4}

AddTerms

Add terms

m = \frac{3 . 5}{7}

ExpandFrac

$\frac{a}{b} = \frac{a \cdot 2}{b \cdot 2}$

m = \frac{7}{1 4}

ReduceFrac

$\frac{a}{b} = \frac{a / 7}{b / 7}$

m = \frac{1}{2}

The slope of the line that passes through the stars is

\frac{1}{2},

0.5 .

Parallel lines have the same slope. Therefore, all lines that are parallel to the line that passes through the stars will have a slope of

\frac{1}{2} .

A general equation in slope-intercept form for these lines can be written.

y = \frac{1}{2} x + b

The parallel line should pass through the moon, which in the coordinate plane has coordinates

(- 2, 1) .

These two numbers will be substituted into the above equation. Then, the

y -

intercept

b

of the parallel line can be determined.

y = \frac{1}{2} x + b

SubstituteII

$x = - 2$ , $y = 1$

1 = \frac{1}{2} (- 2) + b

▼

Solve for

b

MoveRightFacToNumOne

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

1 = \frac{- 2}{2} + b

MoveNegNumToFrac

Put minus sign in front of fraction

1 = - \frac{2}{2} + b

QuotOne

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

1 = - 1 + b

AddEqn

$LHS + 1 = RHS + 1$

2 = b

RearrangeEqn

Rearrange equation

b = 2

Now, the equation of the parallel line can be completed.

y = \frac{1}{2} x + 2

The solution will be verified by graphing the above line on the visual representation of Zosia's proposed mural.

Explore

Observing Perpendicular Lines

The following applet shows two perpendicular lines and their equations in slope-intercept form.

The lines remain perpendicular as they are moved. What characteristic of the lines does not change? What is the relation between the slopes of these two perpendicular lines?

Discussion

Perpendicular Lines

Two coplanar lines — lines that are on the same plane — that intersect at a right angle are said to be perpendicular lines. The symbol $⊥$ is used to algebraically denote that two lines are perpendicular. In the diagram, lines $m$ and $ℓ$ 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

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

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Catch-Up and Review

Proof

Part 1

Part 2

Hint

Solution

Hint

Solution

Hint

Solution

Proof

Part 1

Part $1$

Part $2$

Part $1$