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

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

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}

MoveNegDenomToFrac

Put minus sign in front of fraction

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

QuotOne

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

- 1 = m_{1} m_{2}

RearrangeEqn

Rearrange equation

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

It has been proven that if two lines are perpendicular, then the product of their slopes is

- 1 .

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

Part $2$

Here it is assumed that the slopes of two lines

ℓ_{1}

and

ℓ_{2}

are opposite reciprocals.

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

Consider the steps taken in Part

1 .

This time, it should be found that

△ A O C

is a right triangle.

If the lengths of the sides of

△ A O C

satisfy the Pythagorean Theorem, then the triangle is a right triangle.

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

The side lengths, which were previously found in Part

1,

can be substituted into the above equation.

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}

Substitute

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

2 = ? - 2 (- 1)

MultNegNegOnePar

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

2 = 2 ✓

Since a true statement was obtained,

△ A O C

is a right triangle. Therefore,

ℓ_{1}

and

ℓ_{2}

are perpendicular lines. This completes the second part.

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

The biconditional statement has been proven.

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

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

Finding a Perpendicular Line Through a Point

Write an equation in slope-intercept form of the line that passes through $(6, - 1)$ and is perpendicular to $y = - 3 x + 2 .$

Hint

What makes two lines perpendicular?

Solution

Consider the given equation.

y = - 3 x + 2

Two lines are perpendicular if and only if their slopes are negative reciprocals. This means that their product is

- 1 .

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

The slope of the given equation is

- 3 .

By substituting this value into the above equation the slope of a perpendicular line

m_{2}

can be found.

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

Substitute

$m_{1} = - 3$

- 3 \cdot m_{2} = - 1

▼

Solve for

m_{2}

DivEqn

$LHS / (- 3) = RHS / (- 3)$

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

DivNegNeg

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

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

All perpendicular lines to the line whose equation is given have a slope of

\frac{1}{3} .

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

y = \frac{1}{3} x + b

The desired perpendicular line passes through the point with coordinates

(6, - 1) .

By substituting these numbers into the obtained equation for

x

and

y,

the

y -

intercept

b

of the perpendicular line can be determined.

y = \frac{1}{3} x + b

SubstituteII

$x = 6$ , $y = - 1$

- 1 = \frac{1}{3} (6) + b

▼

Solve for

b

MoveRightFacToNumOne

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

- 1 = \frac{6}{3} + b

CalcQuot

Calculate quotient

- 1 = 2 + b

SubEqn

$LHS - 2 = RHS - 2$

- 3 = b

RearrangeEqn

Rearrange equation

b = - 3

Knowing the

y -

intercept, the equation of the perpendicular line to

y = - 3 x + 2

through

(6, - 1)

can now be written.

y = \frac{1}{3} x + (- 3) \Leftrightarrow y = \frac{1}{3} x - 3

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

As seen above, the graph of $y = \frac{1}{3} x - 3$ is perpendicular to the graph of $y = - 3 x + 2$ and passes through $(6, - 1) .$

Example

Water Supply Network

Zain's class is modeling a neighborhood that is being built outside of town. For extra credit, Zain decides to use the neighborhood's plumbing plan determine where the pipe that connects a new house to the water supply network will be placed. They have the following plan of the network.

Zain knows that the new pipe has to be perpendicular to the main pipe. It also must pass through

(3, - 1),

which represents the new house. Help Zain find the equation in slope-intercept of the line that represents the new pipe.

Hint

Start by finding the slope of the line that represents the main pipe.

Solution

First, the slope of the line that represents the main pipe will be determined. To do so, the Slope Formula will be used.

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

Here,

(x_{1}, y_{1})

and

(x_{2}, y_{2})

are the coordinates of two points on the line. Therefore, two points on the given line are needed.

The coordinates of two points on the main pipe are

(0, 2)

and

(2, 3) .

These will be substituted into the formula.

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

SubstitutePoints

Substitute $(0, 2)$ & $(2, 3)$

m = \frac{3 - 2}{2 - 0}

SubTerms

Subtract terms

m = \frac{1}{2}

The slope of the line that represents the main pipe is

\frac{1}{2} .

Two lines are perpendicular when their slopes are negative reciprocals. This means that the product of the slopes is

- 1 .

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

The slope of the given line is

\frac{1}{2} .

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

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

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

MultEqn

$LHS \cdot 2 = RHS \cdot 2$

m_{2} = - 2

All lines perpendicular to the line that represents the main pipe 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 corresponds to the new pipe should pass through

(3, - 1) .

By substituting

3

and

- 1

in the general equation for

x

and

y,

the

y -

intercept

b

of the perpendicular line can be determined.

y = - 2 x + b

SubstituteII

$x = 3$ , $y = - 1$

- 1 = - 2 (3) + b

▼

Solve for

b

MultNegPos

$(- a) b = - a b$

- 1 = - 6 + b

AddEqn

$LHS + 6 = RHS + 6$

5 = b

RearrangeEqn

Rearrange equation

b = 5

Knowing the

y -

intercept, the equation of the new pipe can now be written.

y = - 2 x + 5

The solution can be verified by graphing the line on the given plan.

As seen above, the graph of $y = - 2 x + 5$ is perpendicular to the given line and passes through $(3, - 1) .$ The new pipe is a part of $y = - 2 x + 5 .$

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.

Closure

Lines in the Three-Dimensional Space

As mentioned at the beginning of the lesson, two lines in a plane are either parallel or intersecting lines. A particular type of intersecting lines are be perpendicular lines.

Examples of parallel and intersecting lines.

In a three-dimensional space, there is a third option — skew lines. Skew lines are neither parallel nor intersecting because they lie on different planes.

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Catch-Up and Review

Proof

Part 1

Part 2

Hint

Solution

Hint

Solution

Hint

Solution

Proof

Part 1

Part 2

Hint

Solution

Hint

Solution

Hint

Solution

Part $1$

Part $2$

Part $1$

Part $2$