Understanding Parallel and Perpendicular Lines: Equations Explored

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?

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.

Pop Quiz

Identifying Perpendicular Lines

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

Example

Space Planning

Tiffaniqua is designing a new community garden for the class proposal. The entrance to the garden is in its corner. In the opposite corner there is a wooden bench. She plans to place flower beds in another corner of the garden. Consider a plan of the garden made in a specialized software.

Tiffaniqua wants to have a stone path connecting the bench with the entrance. She also wants another stone path that would allow visitors to easily access the flower beds. The paths should be perpendicular to each other. What equation, written in slope-intercept form, represents the path that will let the community garden visitors access the flowers?

Hint

Start by finding the slope of the line that connects the bench and the entrance.

Solution

The stone path that connects the bench with the entrance lies on the line that passes through the points with coordinates

(0, 4)

and

(6, 0) .

These can be used to find the line's slope by using the Slope Formula.

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

SubstitutePoints

Substitute $(0, 4)$ & $(6, 0)$

m = \frac{0 - 4}{6 - 0}

Evaluate right-hand side

SubTerms

Subtract terms

m = \frac{- 4}{6}

MoveNegNumToFrac

Put minus sign in front of fraction

m = - \frac{4}{6}

ReduceFrac

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

m = - \frac{2}{3}

The

x -

coordinate of

(0, 4)

0 .

Therefore, the

y -

intercept of the line through this point is

4 .

Knowing the slope an the

y -

intercept of the line, its equation in slope-intercept form can be written.

y = - \frac{2}{3} x + 4

The path from the bench to the entrance is a part of the line described by the above equation. Note that the whole equation is not actually needed. It would be enough to know only the slope of the line passing through the bench and the entrance to find a line perpendicular to it.

Now, the equation of the line that corresponds to the second path must be found. First, the slope of this line will be determined. The second line is perpendicular to

y = - \frac{2}{3} x + 4 .

Perpendicular lines have negative reciprocal slopes. This means that their product is

- 1 .

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

The slope of the known equation is

- \frac{2}{3} .

By substituting this value into the above equation for

m_{1},

the slope of a perpendicular line

m_{2}

can be found.

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

Substitute

$m_{1} = - \frac{2}{3}$

- \frac{2}{3} \cdot m_{2} = - 1

Solve for

m_{2}

MultEqn

$LHS \cdot 3 = RHS \cdot 3$

- 2 \cdot m_{2} = - 3

DivEqn

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

m_{2} = \frac{- 3}{- 2}

DivNegNeg

$\frac{- a}{- b} = \frac{a}{b}$

m_{2} = \frac{3}{2}

All lines perpendicular to the line that passes through the bench and the entrance have a slope of

\frac{3}{2} .

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

y = \frac{3}{2} x + b

The desired line has to allow community members to easily access the flower beds. This means that the line should pass through the point with coordinates

(6, 4) .

By substituting them into the above equation for

x

and

y,

the

y -

intercept

b

of the perpendicular line can be found.

y = \frac{3}{2} x + b

SubstituteII

$x = 6$ , $y = 4$

4 = \frac{3}{2} (6) + b

Solve for

b

MoveRightFacToNum

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

4 = \frac{1 8}{2} + b

CalcQuot

Calculate quotient

4 = 9 + b

SubEqn

$LHS - 9 = RHS - 9$

- 5 = b

RearrangeEqn

Rearrange equation

b = - 5

Knowing the

y -

intercept, the equation of the line perpendicular to

y = - \frac{2}{3} x + 4

through

(6, 4)

can be written.

y = \frac{3}{2} x + (- 5) \Leftrightarrow y = \frac{3}{2} x - 5

The path that will let visitors to the garden access the flowers is a part of the above line. Below it is shown how the solution looks in the specialized software.

It can be seen that the lines are perpendicular and that $y = \frac{3}{2} x - 5$ passes through $(6, 4),$ which corresponds to the flower beds.

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Proof

Part 1

Part 2

Hint

Solution

Part $1$

Part $2$