Parallel and Perpendicular Lines

Introduction
Identifying Parallel Lines
Identifying Perpendicular Lines
Forms of Line Equations
Finding Equations of Parallel & Perpendicular Lines
Problem Examples
Real-world Applications
Test Your Knowledge: Quiz

Introduction to Parallel and Perpendicular Lines

In coordinate geometry, understanding the relationship between lines is fundamental. Two of the most important relationships are:

Parallel Lines: Two lines that never intersect and maintain the same distance between them.

Perpendicular Lines: Two lines that intersect at a 90° angle (right angle).

Key Point: The relationship between lines can be determined algebraically using their slopes or equations, without having to graph them.

Identifying Parallel Lines

There are several ways to identify whether two lines are parallel:

Using Slope

Two lines with slopes m₁ and m₂ are parallel if and only if:
m₁ = m₂

Example 1: Determining if lines are parallel using slopes

Are the lines with the following equations parallel?

Line 1: y = 2x + 3
Line 2: y = 2x - 5

Solution

To determine if these lines are parallel, we need to compare their slopes.

Line 1: y = 2x + 3, has slope m₁ = 2
Line 2: y = 2x - 5, has slope m₂ = 2

Since m₁ = m₂, the lines are parallel.

Using General Form

For lines in general form Ax + By + C = 0, two lines are parallel if their coefficients A and B are proportional:
A₁/A₂ = B₁/B₂

Example 2: Determining if lines are parallel using general form

Are the following lines parallel?

Line 1: 3x - 6y + 12 = 0
Line 2: 2x - 4y + 7 = 0

Solution

Line 1: 3x - 6y + 12 = 0, has coefficients A₁ = 3, B₁ = -6
Line 2: 2x - 4y + 7 = 0, has coefficients A₂ = 2, B₂ = -4

Let's check if A₁/A₂ = B₁/B₂:
A₁/A₂ = 3/2 = 1.5
B₁/B₂ = (-6)/(-4) = 1.5

Since A₁/A₂ = B₁/B₂, the lines are parallel.

Identifying Perpendicular Lines

Using Slope

Two lines with slopes m₁ and m₂ are perpendicular if and only if:
m₁ × m₂ = -1
(Their slopes are negative reciprocals of each other)

Example 3: Determining if lines are perpendicular using slopes

Are the lines with the following equations perpendicular?

Line 1: y = 2x + 3
Line 2: y = -1/2x + 4

Solution

To determine if these lines are perpendicular, we check if their slopes multiply to give -1.

Line 1: y = 2x + 3, has slope m₁ = 2
Line 2: y = -1/2x + 4, has slope m₂ = -1/2

m₁ × m₂ = 2 × (-1/2) = -1

Since m₁ × m₂ = -1, the lines are perpendicular.

Using General Form

For lines in general form A₁x + B₁y + C₁ = 0 and A₂x + B₂y + C₂ = 0, the lines are perpendicular if:
A₁A₂ + B₁B₂ = 0

Example 4: Determining if lines are perpendicular using general form

Are the following lines perpendicular?

Line 1: 3x + 4y - 10 = 0
Line 2: 4x - 3y + 5 = 0

Solution

Line 1: 3x + 4y - 10 = 0, has coefficients A₁ = 3, B₁ = 4
Line 2: 4x - 3y + 5 = 0, has coefficients A₂ = 4, B₂ = -3

A₁A₂ + B₁B₂ = (3 × 4) + (4 × (-3)) = 12 - 12 = 0

Since A₁A₂ + B₁B₂ = 0, the lines are perpendicular.

Remember: Vertical lines (x = a) have undefined slope, and horizontal lines (y = b) have zero slope. A vertical line is perpendicular to a horizontal line, but two vertical lines are parallel to each other, as are two horizontal lines.

Forms of Line Equations

Before diving deeper, let's review the different forms of line equations:

Form	Equation	Description
Slope-Intercept	y = mx + b	m is the slope, b is the y-intercept
Point-Slope	y - y₁ = m(x - x₁)	m is the slope, (x₁, y₁) is a point on the line
General/Standard	Ax + By + C = 0	A, B, C are constants, A and B not both zero
Intercept	x/a + y/b = 1	a is the x-intercept, b is the y-intercept

Converting between forms: Being able to convert between these forms is essential for working with parallel and perpendicular lines.

Finding Equations of Parallel & Perpendicular Lines

Finding Equation of a Parallel Line

To find the equation of a line parallel to y = mx + b passing through point (x₁, y₁):
1. Use the same slope m
2. Use point-slope form: y - y₁ = m(x - x₁)
3. Simplify to your desired form

Example 5: Finding the equation of a parallel line

Find the equation of the line that is parallel to y = 3x - 4 and passes through the point (2, 5).

Solution

The original line y = 3x - 4 has slope m = 3.

Since parallel lines have the same slope, our new line will also have slope m = 3.

Using the point-slope form with point (2, 5):

y - 5 = 3(x - 2)

Expanding: y - 5 = 3x - 6

Solving for y: y = 3x - 1

Therefore, the equation of the parallel line is y = 3x - 1.

Finding Equation of a Perpendicular Line

To find the equation of a line perpendicular to y = mx + b passing through point (x₁, y₁):
1. Calculate the perpendicular slope: m_perp = -1/m
2. Use point-slope form: y - y₁ = m_perp(x - x₁)
3. Simplify to your desired form

Example 6: Finding the equation of a perpendicular line

Find the equation of the line that is perpendicular to y = 2x + 3 and passes through the point (4, -1).

Solution

The original line y = 2x + 3 has slope m = 2.

The perpendicular slope will be m_perp = -1/m = -1/2.

Using the point-slope form with point (4, -1):

y - (-1) = (-1/2)(x - 4)

Simplifying: y + 1 = (-1/2)x + 2

Solving for y: y = -1/2x + 1

Therefore, the equation of the perpendicular line is y = -1/2x + 1.

Problem Examples

Example 7: Determining the relationship between two lines

Determine whether the following pairs of lines are parallel, perpendicular, or neither:

a) 2x - 3y + 4 = 0 and 4x - 6y - 5 = 0
b) 5x + 2y - 7 = 0 and 2x - 5y + 8 = 0

Solution

a) 2x - 3y + 4 = 0 and 4x - 6y - 5 = 0

Rearranging to slope-intercept form:

First line: 2x - 3y + 4 = 0 → -3y = -2x - 4 → y = (2/3)x + (4/3)

Second line: 4x - 6y - 5 = 0 → -6y = -4x + 5 → y = (2/3)x - (5/6)

The slopes are both 2/3, so the lines are parallel.

b) 5x + 2y - 7 = 0 and 2x - 5y + 8 = 0

Rearranging to slope-intercept form:

First line: 5x + 2y - 7 = 0 → 2y = -5x + 7 → y = (-5/2)x + (7/2)

Second line: 2x - 5y + 8 = 0 → -5y = -2x - 8 → y = (2/5)x + (8/5)

Let's check if these slopes are negative reciprocals:

m₁ × m₂ = (-5/2) × (2/5) = -1

Since m₁ × m₂ = -1, the lines are perpendicular.

Example 8: Distance from a point to a line

Find the distance from the point (3, 4) to the line 2x + 3y - 12 = 0.

Solution

To find the distance from a point to a line, we use the formula:

d = |Ax₀ + By₀ + C| / √(A² + B²)

Where (x₀, y₀) is the point, and Ax + By + C = 0 is the line equation.

For our problem:

A = 2, B = 3, C = -12, (x₀, y₀) = (3, 4)

d = |2(3) + 3(4) - 12| / √(2² + 3²) = |6 + 12 - 12| / √13 = |6| / √13 = 6/√13 ≈ 1.66

Therefore, the distance from point (3, 4) to the line 2x + 3y - 12 = 0 is 6/√13 units.

Example 9: Finding the perpendicular bisector

Find the equation of the perpendicular bisector of the line segment joining the points A(1, 3) and B(5, 7).

Solution

Step 1: Find the midpoint of AB.

Midpoint = ((x₁ + x₂)/2, (y₁ + y₂)/2) = ((1 + 5)/2, (3 + 7)/2) = (3, 5)

Step 2: Find the slope of AB.

Slope of AB = (y₂ - y₁)/(x₂ - x₁) = (7 - 3)/(5 - 1) = 4/4 = 1

Step 3: Find the slope of the perpendicular bisector.

Perpendicular slope = -1/1 = -1

Step 4: Write the equation of the perpendicular bisector using the point-slope form.

Using the point (3, 5) and slope -1:

y - 5 = -1(x - 3)

y - 5 = -x + 3

y = -x + 8

Therefore, the equation of the perpendicular bisector is y = -x + 8.

Real-world Applications

Parallel and perpendicular lines have numerous real-world applications:

Architecture: Design of buildings, where perpendicular walls create right angles for stability and parallel floors maintain consistent spacing.
Engineering: Design of road systems, railroad tracks, and electrical circuits often involve parallel and perpendicular lines.
Computer Graphics: Creating and manipulating shapes and images using coordinate geometry.
Navigation: Latitude and longitude lines on maps form a grid of perpendicular lines.
Physics: Force vectors in mechanics often use perpendicular components.

Example 10: Construction Application

A rectangular building is planned with one side along the line y = 2x + 5. The adjacent side needs to be perpendicular to this line. Find the equation of the line representing this adjacent side if it passes through the point (3, 11).

Solution

The first side of the building follows the line y = 2x + 5, which has a slope of 2.

For the adjacent side to be perpendicular, its slope must be the negative reciprocal: -1/2.

Using point-slope form with point (3, 11):

y - 11 = (-1/2)(x - 3)

Expanding: y - 11 = -1/2x + 3/2

Solving for y: y = -1/2x + 25/2

Therefore, the equation of the adjacent side is y = -1/2x + 12.5.