Learn on PengiSaxon Algebra 1Chapter 6: Polynomials and Factoring

Lesson 52: Determining the Equation of a Line Given Two Points

In this Grade 9 Saxon Algebra 1 lesson, students learn how to determine the equation of a line using the point-slope form (y − y₁ = m(x − x₁)), given either a slope and a single point or two coordinate points. Students practice calculating slope using the slope formula, converting point-slope form to slope-intercept form, and applying these skills to real-world problems. This lesson is part of Chapter 6 on Polynomials and Factoring and builds on prior knowledge of slope, intercepts, and graphing lines.

Section 1

📘 Determining the Equation of a Line Given Two Points

New Concept

The form yy1=m(xx1)y - y_1 = m(x - x_1), where mm is the slope and (x1,y1)(x_1, y_1) is a point on the line, is called the point-slope form.

What’s next

Next, you’ll use this form to write equations for lines when given any two points, applying it to solve problems.

Section 2

Point-Slope Form

Property

The form yy1=m(xx1)y - y_1 = m(x - x_1), where mm is the slope and (x1,y1)(x_1, y_1) is a point on the line is called the point-slope form of a line.

Examples

Write the equation for a line with slope 3 passing through (2, 4): y4=3(x2)y - 4 = 3(x - 2).
Write the equation for a line with slope -5 passing through (1, 8): y8=5(x1)y - 8 = -5(x - 1).
Write the equation for a line with slope 12\frac{1}{2} passing through (-3, 6): y6=12(x+3)y - 6 = \frac{1}{2}(x + 3).

Explanation

Think of this as your instant recipe for a line! If you have one point and the slope (the 'steepness'), you can plug them straight into this formula. It is the fastest way to write the equation without finding the y-intercept first. Super handy for quick equations!

Section 3

Writing an Equation from Two Points

Property

First, find the slope with m=y2y1x2x1m = \frac{y_2 - y_1}{x_2 - x_1}. Then, use the point-slope form yy1=m(xx1)y - y_1 = m(x - x_1) to write the equation.

Examples

Find the equation for a line through (2, 3) and (5, 9): First, m=9352=63=2m = \frac{9 - 3}{5 - 2} = \frac{6}{3} = 2. Then, use a point: y9=2(x5)y - 9 = 2(x - 5).
Find the equation for a line through (1, -3) and (4, 5): First, m=5(3)41=83m = \frac{5 - (-3)}{4 - 1} = \frac{8}{3}. Then, use a point: y5=83(x4)y - 5 = \frac{8}{3}(x - 4).

Explanation

Got two points but no slope? No problem! First, calculate the 'rise over run' between the two points to find your slope. Then, just pick one of those points, grab your new slope, and pop them right into the point-slope formula. You've officially built a line from scratch!

Section 4

Graphing with a Point and Slope

Property

To graph a line with a known slope and point, first plot the given point. Then, use the slope (rise over run) from that point to find a second point and draw a line through both.

Examples

Graph a line with slope 4 and point (3, 5): Plot (3, 5). From there, go up 4 and right 1 to find point (4, 9). Draw the line.
Graph a line with slope 23-\frac{2}{3} and point (2, 4): Plot (2, 4). From there, go down 2 and right 3 to find point (5, 2). Draw the line.

Explanation

It is like a treasure map! Start at your given point ('X' marks the spot). The slope tells you where to go next: 'rise' is your up/down count, and 'run' is your right count. Plot your second point where you land and connect the dots to reveal the treasure path!

Book overview

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

Continue this chapter

Chapter 6: Polynomials and Factoring

  1. Lesson 1

    Lesson 51: Simplifying Rational Expressions with Like Denominators

  2. Lesson 2Current

    Lesson 52: Determining the Equation of a Line Given Two Points

  3. Lesson 3

    Lesson 53: Adding and Subtracting Polynomials

  4. Lesson 4

    Lesson 54: Displaying Data in a Box-and-Whisker Plot

  5. Lesson 5

    Lesson 55: Solving Systems of Linear Equations by Graphing

  6. Lesson 6

    Lesson 56: Identifying, Writing, and Graphing Direct Variation

  7. Lesson 7

    Lesson 57: Finding the Least Common Multiple

  8. Lesson 8

    Lesson 58: Multiplying Polynomials

  9. Lesson 9

    Lesson 59: Solving Systems of Linear Equations by Substitution

  10. Lesson 10

    Lesson 60: Finding Special Products of Binomials

Lesson overview

Expand to review the lesson summary and core properties.

Expand

Section 1

📘 Determining the Equation of a Line Given Two Points

New Concept

The form yy1=m(xx1)y - y_1 = m(x - x_1), where mm is the slope and (x1,y1)(x_1, y_1) is a point on the line, is called the point-slope form.

What’s next

Next, you’ll use this form to write equations for lines when given any two points, applying it to solve problems.

Section 2

Point-Slope Form

Property

The form yy1=m(xx1)y - y_1 = m(x - x_1), where mm is the slope and (x1,y1)(x_1, y_1) is a point on the line is called the point-slope form of a line.

Examples

Write the equation for a line with slope 3 passing through (2, 4): y4=3(x2)y - 4 = 3(x - 2).
Write the equation for a line with slope -5 passing through (1, 8): y8=5(x1)y - 8 = -5(x - 1).
Write the equation for a line with slope 12\frac{1}{2} passing through (-3, 6): y6=12(x+3)y - 6 = \frac{1}{2}(x + 3).

Explanation

Think of this as your instant recipe for a line! If you have one point and the slope (the 'steepness'), you can plug them straight into this formula. It is the fastest way to write the equation without finding the y-intercept first. Super handy for quick equations!

Section 3

Writing an Equation from Two Points

Property

First, find the slope with m=y2y1x2x1m = \frac{y_2 - y_1}{x_2 - x_1}. Then, use the point-slope form yy1=m(xx1)y - y_1 = m(x - x_1) to write the equation.

Examples

Find the equation for a line through (2, 3) and (5, 9): First, m=9352=63=2m = \frac{9 - 3}{5 - 2} = \frac{6}{3} = 2. Then, use a point: y9=2(x5)y - 9 = 2(x - 5).
Find the equation for a line through (1, -3) and (4, 5): First, m=5(3)41=83m = \frac{5 - (-3)}{4 - 1} = \frac{8}{3}. Then, use a point: y5=83(x4)y - 5 = \frac{8}{3}(x - 4).

Explanation

Got two points but no slope? No problem! First, calculate the 'rise over run' between the two points to find your slope. Then, just pick one of those points, grab your new slope, and pop them right into the point-slope formula. You've officially built a line from scratch!

Section 4

Graphing with a Point and Slope

Property

To graph a line with a known slope and point, first plot the given point. Then, use the slope (rise over run) from that point to find a second point and draw a line through both.

Examples

Graph a line with slope 4 and point (3, 5): Plot (3, 5). From there, go up 4 and right 1 to find point (4, 9). Draw the line.
Graph a line with slope 23-\frac{2}{3} and point (2, 4): Plot (2, 4). From there, go down 2 and right 3 to find point (5, 2). Draw the line.

Explanation

It is like a treasure map! Start at your given point ('X' marks the spot). The slope tells you where to go next: 'rise' is your up/down count, and 'run' is your right count. Plot your second point where you land and connect the dots to reveal the treasure path!

Book overview

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

Continue this chapter

Chapter 6: Polynomials and Factoring

  1. Lesson 1

    Lesson 51: Simplifying Rational Expressions with Like Denominators

  2. Lesson 2Current

    Lesson 52: Determining the Equation of a Line Given Two Points

  3. Lesson 3

    Lesson 53: Adding and Subtracting Polynomials

  4. Lesson 4

    Lesson 54: Displaying Data in a Box-and-Whisker Plot

  5. Lesson 5

    Lesson 55: Solving Systems of Linear Equations by Graphing

  6. Lesson 6

    Lesson 56: Identifying, Writing, and Graphing Direct Variation

  7. Lesson 7

    Lesson 57: Finding the Least Common Multiple

  8. Lesson 8

    Lesson 58: Multiplying Polynomials

  9. Lesson 9

    Lesson 59: Solving Systems of Linear Equations by Substitution

  10. Lesson 10

    Lesson 60: Finding Special Products of Binomials