Learn on PengiSaxon Algebra 1Chapter 7: Rational Expressions and Radicals

Lesson 65: Writing Equations of Parallel and Perpendicular Lines

In this Grade 9 Saxon Algebra 1 lesson, students learn how to identify and write equations of parallel and perpendicular lines using slope-intercept form and the point-slope formula. The lesson covers the key slope relationships that define these lines — parallel lines share the same slope, while perpendicular lines have slopes that are negative reciprocals of each other. Students practice converting equations, applying these slope rules, and use coordinate geometry to classify triangles by analyzing line segments.

Section 1

📘 Writing Equations of Parallel and Perpendicular Lines

New Concept

Two nonvertical lines are parallel if they have the same slope and are not the same line.

What’s next

Next, you’ll use the rules for parallel and perpendicular slopes to write new linear equations and analyze geometric figures on the coordinate plane.

Section 2

Slopes of Parallel Lines

Property

Two nonvertical lines are parallel if they have the same slope and are not the same line. Any two vertical lines are parallel.

Explanation

Think of parallel lines as perfect twin roads running side-by-side. They never cross because they have the exact same steepness, or slope. The only difference is their starting point, or y-intercept. If they had the same slope and intercept, they would be the exact same line! So, for lines to be parallel, remember: same slope, different y-intercept.

Examples

The lines y=5x+8y = 5x + 8 and y=5x2y = 5x - 2 are parallel because they both have a slope of 55.
The equations y=13x+2y = -\frac{1}{3}x + 2 and x+3y=9x + 3y = 9 are parallel because solving the second for yy gives y=13x+3y = -\frac{1}{3}x + 3, showing they share the same slope.

Section 3

Example Card: Writing Equations of Parallel Lines

Let's use one line as a blueprint to build another that runs perfectly alongside it. This example tackles the first key idea: writing equations for parallel lines.

Example Problem

Write an equation in slope-intercept form for the line that passes through (-3, 5) and is parallel to a line with equation y=3x+2y = 3x + 2.

Step-by-Step

  1. First, determine the slope of the given line. The equation y=3x+2y = 3x + 2 is in slope-intercept form, so its slope is 33. Any line parallel to it must have the same slope.
  2. Now we have the slope m=3m = 3 and a point (-3, 5). We can use the point-slope formula to find the new line's equation.
yy1=m(xx1) y - y_1 = m(x - x_1)
  1. Substitute the slope and the coordinates of the point into the formula.
y5=3(x(3)) y - 5 = 3(x - (-3))
  1. Simplify the expression inside the parentheses.
y5=3(x+3) y - 5 = 3(x + 3)
  1. Apply the Distributive Property to the right side of the equation.
y5=3x+9 y - 5 = 3x + 9
  1. To get the final equation in slope-intercept form, isolate yy by adding 55 to both sides.
y=3x+14 y = 3x + 14

Section 4

Slopes of Perpendicular Lines

Property

Any two lines are perpendicular if their slopes are negative reciprocals of each other. A vertical and horizontal line are also perpendicular.

Explanation

Perpendicular lines are like two paths that intersect to form a perfect 'T' or a 90-degree angle. For this to happen, their slopes must be mathematical opposites. You take the slope of one line, flip it upside down (that's the reciprocal), and then reverse its sign (that's the negative). This special relationship guarantees a perfect right-angle intersection every time.

Examples

The line y=2x+1y = 2x + 1 is perpendicular to y=12x4y = -\frac{1}{2}x - 4 because their slopes, 22 and 12-\frac{1}{2}, are negative reciprocals.
A line with slope 34-\frac{3}{4} is perpendicular to a line with slope 43\frac{4}{3}, since (34)(43)=1(-\frac{3}{4}) \cdot (\frac{4}{3}) = -1.
A vertical line like x=5x = 5 is perpendicular to a horizontal line like y=2y = -2.

Section 5

Example Card: Writing Equations of Perpendicular Lines

Ready to make a sharp 90-degree turn? This example shows how to build a line that's perfectly perpendicular to another, focusing on the second key idea.

Example Problem

Write an equation in slope-intercept form for the line that passes through (4, -1) and is perpendicular to a line with equation y=2x+5y = -2x + 5.

Step-by-Step

  1. First, find the slope of the given line, y=2x+5y = -2x + 5. The slope is 2-2.
  2. A perpendicular line has a slope that is the negative reciprocal of the original slope. The negative reciprocal of 2-2 is 12\frac{1}{2}. So, for our new line, m=12m = \frac{1}{2}.
  3. Use the point-slope formula, yy1=m(xx1)y - y_1 = m(x - x_1), with the new slope and the given point (4, -1).
  4. Substitute the values into the formula.
y(1)=12(x4) y - (-1) = \frac{1}{2}(x - 4)
  1. Simplify the equation.
y+1=12(x4) y + 1 = \frac{1}{2}(x - 4)
  1. Apply the Distributive Property to the right side.
y+1=12x2 y + 1 = \frac{1}{2}x - 2
  1. Isolate yy by subtracting 11 from both sides to get the slope-intercept form.
y=12x3 y = \frac{1}{2}x - 3

Book overview

Jump across lessons in the current chapter without opening the full course modal.

Continue this chapter

Chapter 7: Rational Expressions and Radicals

  1. Lesson 1

    Lesson 61: Simplifying Radical Expressions

  2. Lesson 2

    Lesson 62: Displaying Data in Stem-and-Leaf Plots and Histograms

  3. Lesson 3

    Lesson 63: Solving Systems of Linear Equations by Elimination

  4. Lesson 4

    Lesson 64: Identifying, Writing, and Graphing Inverse Variation

  5. Lesson 5Current

    Lesson 65: Writing Equations of Parallel and Perpendicular Lines

  6. Lesson 6

    Lesson 66: Solving Inequalities by Adding or Subtracting

  7. Lesson 7

    Lesson 67: Solving and Classifying Special Systems of Linear Equations

  8. Lesson 8

    Lesson 68: Mutually Exclusive and Inclusive Events

  9. Lesson 9

    Lesson 69: Adding and Subtracting Radical Expressions

  10. Lesson 10

    Lesson 70: Solving Inequalities by Multiplying or Dividing

Lesson overview

Expand to review the lesson summary and core properties.

Expand

Section 1

📘 Writing Equations of Parallel and Perpendicular Lines

New Concept

Two nonvertical lines are parallel if they have the same slope and are not the same line.

What’s next

Next, you’ll use the rules for parallel and perpendicular slopes to write new linear equations and analyze geometric figures on the coordinate plane.

Section 2

Slopes of Parallel Lines

Property

Two nonvertical lines are parallel if they have the same slope and are not the same line. Any two vertical lines are parallel.

Explanation

Think of parallel lines as perfect twin roads running side-by-side. They never cross because they have the exact same steepness, or slope. The only difference is their starting point, or y-intercept. If they had the same slope and intercept, they would be the exact same line! So, for lines to be parallel, remember: same slope, different y-intercept.

Examples

The lines y=5x+8y = 5x + 8 and y=5x2y = 5x - 2 are parallel because they both have a slope of 55.
The equations y=13x+2y = -\frac{1}{3}x + 2 and x+3y=9x + 3y = 9 are parallel because solving the second for yy gives y=13x+3y = -\frac{1}{3}x + 3, showing they share the same slope.

Section 3

Example Card: Writing Equations of Parallel Lines

Let's use one line as a blueprint to build another that runs perfectly alongside it. This example tackles the first key idea: writing equations for parallel lines.

Example Problem

Write an equation in slope-intercept form for the line that passes through (-3, 5) and is parallel to a line with equation y=3x+2y = 3x + 2.

Step-by-Step

  1. First, determine the slope of the given line. The equation y=3x+2y = 3x + 2 is in slope-intercept form, so its slope is 33. Any line parallel to it must have the same slope.
  2. Now we have the slope m=3m = 3 and a point (-3, 5). We can use the point-slope formula to find the new line's equation.
yy1=m(xx1) y - y_1 = m(x - x_1)
  1. Substitute the slope and the coordinates of the point into the formula.
y5=3(x(3)) y - 5 = 3(x - (-3))
  1. Simplify the expression inside the parentheses.
y5=3(x+3) y - 5 = 3(x + 3)
  1. Apply the Distributive Property to the right side of the equation.
y5=3x+9 y - 5 = 3x + 9
  1. To get the final equation in slope-intercept form, isolate yy by adding 55 to both sides.
y=3x+14 y = 3x + 14

Section 4

Slopes of Perpendicular Lines

Property

Any two lines are perpendicular if their slopes are negative reciprocals of each other. A vertical and horizontal line are also perpendicular.

Explanation

Perpendicular lines are like two paths that intersect to form a perfect 'T' or a 90-degree angle. For this to happen, their slopes must be mathematical opposites. You take the slope of one line, flip it upside down (that's the reciprocal), and then reverse its sign (that's the negative). This special relationship guarantees a perfect right-angle intersection every time.

Examples

The line y=2x+1y = 2x + 1 is perpendicular to y=12x4y = -\frac{1}{2}x - 4 because their slopes, 22 and 12-\frac{1}{2}, are negative reciprocals.
A line with slope 34-\frac{3}{4} is perpendicular to a line with slope 43\frac{4}{3}, since (34)(43)=1(-\frac{3}{4}) \cdot (\frac{4}{3}) = -1.
A vertical line like x=5x = 5 is perpendicular to a horizontal line like y=2y = -2.

Section 5

Example Card: Writing Equations of Perpendicular Lines

Ready to make a sharp 90-degree turn? This example shows how to build a line that's perfectly perpendicular to another, focusing on the second key idea.

Example Problem

Write an equation in slope-intercept form for the line that passes through (4, -1) and is perpendicular to a line with equation y=2x+5y = -2x + 5.

Step-by-Step

  1. First, find the slope of the given line, y=2x+5y = -2x + 5. The slope is 2-2.
  2. A perpendicular line has a slope that is the negative reciprocal of the original slope. The negative reciprocal of 2-2 is 12\frac{1}{2}. So, for our new line, m=12m = \frac{1}{2}.
  3. Use the point-slope formula, yy1=m(xx1)y - y_1 = m(x - x_1), with the new slope and the given point (4, -1).
  4. Substitute the values into the formula.
y(1)=12(x4) y - (-1) = \frac{1}{2}(x - 4)
  1. Simplify the equation.
y+1=12(x4) y + 1 = \frac{1}{2}(x - 4)
  1. Apply the Distributive Property to the right side.
y+1=12x2 y + 1 = \frac{1}{2}x - 2
  1. Isolate yy by subtracting 11 from both sides to get the slope-intercept form.
y=12x3 y = \frac{1}{2}x - 3

Book overview

Jump across lessons in the current chapter without opening the full course modal.

Continue this chapter

Chapter 7: Rational Expressions and Radicals

  1. Lesson 1

    Lesson 61: Simplifying Radical Expressions

  2. Lesson 2

    Lesson 62: Displaying Data in Stem-and-Leaf Plots and Histograms

  3. Lesson 3

    Lesson 63: Solving Systems of Linear Equations by Elimination

  4. Lesson 4

    Lesson 64: Identifying, Writing, and Graphing Inverse Variation

  5. Lesson 5Current

    Lesson 65: Writing Equations of Parallel and Perpendicular Lines

  6. Lesson 6

    Lesson 66: Solving Inequalities by Adding or Subtracting

  7. Lesson 7

    Lesson 67: Solving and Classifying Special Systems of Linear Equations

  8. Lesson 8

    Lesson 68: Mutually Exclusive and Inclusive Events

  9. Lesson 9

    Lesson 69: Adding and Subtracting Radical Expressions

  10. Lesson 10

    Lesson 70: Solving Inequalities by Multiplying or Dividing