Learn on PengiYoshiwara Intermediate AlgebraChapter 1: Linear Models

Lesson 5: Equations of Lines

In this Grade 7 math lesson from Yoshiwara Intermediate Algebra (Chapter 1: Linear Models), students learn to write and interpret equations of lines using slope-intercept form. The lesson applies these concepts to real-world data, including modeling the freezing temperature of seawater as a function of salinity. Students also explore how linear equations connect to climate science topics such as albedo and Arctic sea ice decline.

Section 1

πŸ“˜ Equations of Lines

New Concept

Explore how real-world data, like the effect of salinity on water's freezing point, can be modeled using linear equations. We'll learn to write, graph, and interpret these equations to understand and predict important trends.

What’s next

Next, you’ll work through interactive examples and practice cards to master writing the equation of a line.

Section 2

Slope-Intercept Form

Property

The slope-intercept form for a linear equation is y=mx+by = mx + b, where mm is the slope of the line and the point (0,b)(0, b) is the y-intercept.

The slope tells us the rate of change, and the y-intercept is where the line crosses the vertical axis.

Examples

Find the equation of a line with slope 3 and y-intercept at (0,βˆ’5)(0, -5). Using the slope-intercept form y=mx+by = mx + b, we substitute m=3m=3 and b=βˆ’5b=-5 to get the equation y=3xβˆ’5y = 3x - 5.

Section 3

Point-Slope Form

Property

The point-slope form of a linear equation is y=y1+m(xβˆ’x1)y = y_1 + m(x - x_1), where mm is the slope and (x1,y1)(x_1, y_1) is any known point on the line.

This form is useful for finding the equation of a line when you know its slope and at least one point it passes through.

You may also see the formula written in an alternate version:

yβˆ’y1=m(xβˆ’x1)oryβˆ’y1xβˆ’x1=my-y_1=m(x-x_1) \quad \text{or} \quad \frac{y-y_1}{x-x_1}=m

Section 4

Equation from two points

Property

To find the equation of a line passing through two points, (x1,y1)(x_1, y_1) and (x2,y2)(x_2, y_2), first calculate the slope using the formula m=y2βˆ’y1x2βˆ’x1m = \frac{y_2 - y_1}{x_2 - x_1}. Then, use this slope and one of the points with the point-slope form, yβˆ’y1=m(xβˆ’x1)y - y_1 = m(x - x_1), to write the equation.

Examples

Find the equation of the line passing through (2,5)(2, 5) and (4,9)(4, 9). First, find the slope: m=9βˆ’54βˆ’2=42=2m = \frac{9 - 5}{4 - 2} = \frac{4}{2} = 2. Using point-slope form with (2,5)(2, 5), we get yβˆ’5=2(xβˆ’2)y - 5 = 2(x - 2).

What is the equation of the line containing points (βˆ’1,6)(-1, 6) and (3,βˆ’2)(3, -2)? The slope is m=βˆ’2βˆ’63βˆ’(βˆ’1)=βˆ’84=βˆ’2m = \frac{-2 - 6}{3 - (-1)} = \frac{-8}{4} = -2. Using point (βˆ’1,6)(-1, 6), the equation is yβˆ’6=βˆ’2(x+1)y - 6 = -2(x + 1).

Book overview

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Chapter 1: Linear Models

  1. Lesson 1

    Lesson 1: Linear Models

  2. Lesson 2

    Lesson 2: Graphs and Equations

  3. Lesson 3

    Lesson 3: Intercepts

  4. Lesson 4

    Lesson 4: Slope

  5. Lesson 5Current

    Lesson 5: Equations of Lines

  6. Lesson 6

    Lesson 6: Chapter Summary and Review

Lesson overview

Expand to review the lesson summary and core properties.

Expand

Section 1

πŸ“˜ Equations of Lines

New Concept

Explore how real-world data, like the effect of salinity on water's freezing point, can be modeled using linear equations. We'll learn to write, graph, and interpret these equations to understand and predict important trends.

What’s next

Next, you’ll work through interactive examples and practice cards to master writing the equation of a line.

Section 2

Slope-Intercept Form

Property

The slope-intercept form for a linear equation is y=mx+by = mx + b, where mm is the slope of the line and the point (0,b)(0, b) is the y-intercept.

The slope tells us the rate of change, and the y-intercept is where the line crosses the vertical axis.

Examples

Find the equation of a line with slope 3 and y-intercept at (0,βˆ’5)(0, -5). Using the slope-intercept form y=mx+by = mx + b, we substitute m=3m=3 and b=βˆ’5b=-5 to get the equation y=3xβˆ’5y = 3x - 5.

Section 3

Point-Slope Form

Property

The point-slope form of a linear equation is y=y1+m(xβˆ’x1)y = y_1 + m(x - x_1), where mm is the slope and (x1,y1)(x_1, y_1) is any known point on the line.

This form is useful for finding the equation of a line when you know its slope and at least one point it passes through.

You may also see the formula written in an alternate version:

yβˆ’y1=m(xβˆ’x1)oryβˆ’y1xβˆ’x1=my-y_1=m(x-x_1) \quad \text{or} \quad \frac{y-y_1}{x-x_1}=m

Section 4

Equation from two points

Property

To find the equation of a line passing through two points, (x1,y1)(x_1, y_1) and (x2,y2)(x_2, y_2), first calculate the slope using the formula m=y2βˆ’y1x2βˆ’x1m = \frac{y_2 - y_1}{x_2 - x_1}. Then, use this slope and one of the points with the point-slope form, yβˆ’y1=m(xβˆ’x1)y - y_1 = m(x - x_1), to write the equation.

Examples

Find the equation of the line passing through (2,5)(2, 5) and (4,9)(4, 9). First, find the slope: m=9βˆ’54βˆ’2=42=2m = \frac{9 - 5}{4 - 2} = \frac{4}{2} = 2. Using point-slope form with (2,5)(2, 5), we get yβˆ’5=2(xβˆ’2)y - 5 = 2(x - 2).

What is the equation of the line containing points (βˆ’1,6)(-1, 6) and (3,βˆ’2)(3, -2)? The slope is m=βˆ’2βˆ’63βˆ’(βˆ’1)=βˆ’84=βˆ’2m = \frac{-2 - 6}{3 - (-1)} = \frac{-8}{4} = -2. Using point (βˆ’1,6)(-1, 6), the equation is yβˆ’6=βˆ’2(x+1)y - 6 = -2(x + 1).

Book overview

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

Continue this chapter

Chapter 1: Linear Models

  1. Lesson 1

    Lesson 1: Linear Models

  2. Lesson 2

    Lesson 2: Graphs and Equations

  3. Lesson 3

    Lesson 3: Intercepts

  4. Lesson 4

    Lesson 4: Slope

  5. Lesson 5Current

    Lesson 5: Equations of Lines

  6. Lesson 6

    Lesson 6: Chapter Summary and Review