Learn on PengiAoPS: Introduction to Algebra (AMC 8 & 10)Chapter 8: Graphing Lines

Lesson 2: Introduction to Graphing Linear Equations

In this Grade 4 lesson from AoPS Introduction to Algebra, students learn how to graph two-variable linear equations of the form Ax + By = C on the Cartesian plane and discover why these equations produce straight lines. The lesson introduces slope using the formula m = (y₂ - y₁) / (x₂ - x₁), covering positive, negative, zero, and undefined slope and what each means geometrically. Aligned with AMC 8 and AMC 10 preparation, this lesson builds foundational graphing skills within Chapter 8's study of linear equations.

Section 1

Graph of a Linear Equation

Property

The graph of a linear equation Ax+By=CAx + By = C is a line.

  • Every point on the line is a solution of the equation.
  • Every solution of this equation is a point on this line.

Examples

  • For the equation y=2x+1y = 2x + 1, the point (2,5)(2, 5) is a solution because 5=2(2)+15 = 2(2) + 1. Therefore, the point (2,5)(2, 5) is on the graph of the line.
  • For the same equation, y=2x+1y = 2x + 1, the point (3,6)(3, 6) is not a solution because 62(3)+16 \neq 2(3) + 1. Therefore, the point (3,6)(3, 6) is not on the line.
  • Every point on the graph of x+y=5x + y = 5, such as (1,4)(1, 4) and (5,0)(5, 0), represents a pair of numbers that are solutions to the equation.

Section 2

Graphing by Plotting Points

Property

To graph a linear equation by plotting points:
Step 1. Find three points whose coordinates are solutions to the equation. Organize them in a table.
Step 2. Plot the points in a rectangular coordinate system. Check that the points line up. If they do not, carefully check your work!
Step 3. Draw the line through the three points. Extend the line to fill the grid and put arrows on both ends of the line.

Examples

  • To graph y=x+3y = x + 3, find three solution points like (0,3)(0, 3), (1,4)(1, 4), and (1,2)(-1, 2). Plot these points and draw a line through them.
  • To graph y=12x1y = \frac{1}{2}x - 1, choose multiples of 2 for xx to avoid fractions. Good choices would be (0,1)(0, -1), (2,0)(2, 0), and (4,1)(4, 1).
  • For 3x+y=43x + y = 4, first rewrite it as y=3x+4y = -3x + 4. Then find points by choosing values for xx, such as (0,4)(0, 4), (1,1)(1, 1), and (2,2)(2, -2).

Explanation

This method is like an algebraic connect-the-dots. Find at least three coordinate pairs that solve the equation, place them on the graph, and draw a straight line through them. Using three points helps you catch any calculation mistakes.

Section 3

Slope Formula

Property

The slope of the line between two points (x1,y1)(x_1, y_1) and (x2,y2)(x_2, y_2) is:

m=y2y1x2x1m = \frac{y_2 - y_1}{x_2 - x_1}
This formula calculates the ratio of the change in the y-coordinates (rise) to the change in the x-coordinates (run).

Examples

  • Find the slope between (2,3)(2, 3) and (7,9)(7, 9). Using the formula: m=9372=65m = \frac{9 - 3}{7 - 2} = \frac{6}{5}.
  • Find the slope between (1,5)(-1, 5) and (3,3)(3, -3). Using the formula: m=353(1)=84=2m = \frac{-3 - 5}{3 - (-1)} = \frac{-8}{4} = -2.
  • Find the slope between (4,2)(-4, -2) and (6,8)(-6, 8). Using the formula: m=8(2)6(4)=102=5m = \frac{8 - (-2)}{-6 - (-4)} = \frac{10}{-2} = -5.

Explanation

The slope formula is a way to calculate rise over run without a graph. It finds the vertical distance between points (y2y1)(y_2 - y_1) and divides it by the horizontal distance (x2x1)(x_2 - x_1) to find the steepness.

Book overview

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Chapter 8: Graphing Lines

  1. Lesson 1

    Lesson 1: The Number Line and the Cartesian Plane

  2. Lesson 2Current

    Lesson 2: Introduction to Graphing Linear Equations

  3. Lesson 3

    Lesson 3: Using Slope in Problems

  4. Lesson 4

    Lesson 4: Find the Equation

  5. Lesson 5

    Lesson 5: Slope and Intercepts

  6. Lesson 6

    Lesson 6: Comparing Lines

Lesson overview

Expand to review the lesson summary and core properties.

Expand

Section 1

Graph of a Linear Equation

Property

The graph of a linear equation Ax+By=CAx + By = C is a line.

  • Every point on the line is a solution of the equation.
  • Every solution of this equation is a point on this line.

Examples

  • For the equation y=2x+1y = 2x + 1, the point (2,5)(2, 5) is a solution because 5=2(2)+15 = 2(2) + 1. Therefore, the point (2,5)(2, 5) is on the graph of the line.
  • For the same equation, y=2x+1y = 2x + 1, the point (3,6)(3, 6) is not a solution because 62(3)+16 \neq 2(3) + 1. Therefore, the point (3,6)(3, 6) is not on the line.
  • Every point on the graph of x+y=5x + y = 5, such as (1,4)(1, 4) and (5,0)(5, 0), represents a pair of numbers that are solutions to the equation.

Section 2

Graphing by Plotting Points

Property

To graph a linear equation by plotting points:
Step 1. Find three points whose coordinates are solutions to the equation. Organize them in a table.
Step 2. Plot the points in a rectangular coordinate system. Check that the points line up. If they do not, carefully check your work!
Step 3. Draw the line through the three points. Extend the line to fill the grid and put arrows on both ends of the line.

Examples

  • To graph y=x+3y = x + 3, find three solution points like (0,3)(0, 3), (1,4)(1, 4), and (1,2)(-1, 2). Plot these points and draw a line through them.
  • To graph y=12x1y = \frac{1}{2}x - 1, choose multiples of 2 for xx to avoid fractions. Good choices would be (0,1)(0, -1), (2,0)(2, 0), and (4,1)(4, 1).
  • For 3x+y=43x + y = 4, first rewrite it as y=3x+4y = -3x + 4. Then find points by choosing values for xx, such as (0,4)(0, 4), (1,1)(1, 1), and (2,2)(2, -2).

Explanation

This method is like an algebraic connect-the-dots. Find at least three coordinate pairs that solve the equation, place them on the graph, and draw a straight line through them. Using three points helps you catch any calculation mistakes.

Section 3

Slope Formula

Property

The slope of the line between two points (x1,y1)(x_1, y_1) and (x2,y2)(x_2, y_2) is:

m=y2y1x2x1m = \frac{y_2 - y_1}{x_2 - x_1}
This formula calculates the ratio of the change in the y-coordinates (rise) to the change in the x-coordinates (run).

Examples

  • Find the slope between (2,3)(2, 3) and (7,9)(7, 9). Using the formula: m=9372=65m = \frac{9 - 3}{7 - 2} = \frac{6}{5}.
  • Find the slope between (1,5)(-1, 5) and (3,3)(3, -3). Using the formula: m=353(1)=84=2m = \frac{-3 - 5}{3 - (-1)} = \frac{-8}{4} = -2.
  • Find the slope between (4,2)(-4, -2) and (6,8)(-6, 8). Using the formula: m=8(2)6(4)=102=5m = \frac{8 - (-2)}{-6 - (-4)} = \frac{10}{-2} = -5.

Explanation

The slope formula is a way to calculate rise over run without a graph. It finds the vertical distance between points (y2y1)(y_2 - y_1) and divides it by the horizontal distance (x2x1)(x_2 - x_1) to find the steepness.

Book overview

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

Continue this chapter

Chapter 8: Graphing Lines

  1. Lesson 1

    Lesson 1: The Number Line and the Cartesian Plane

  2. Lesson 2Current

    Lesson 2: Introduction to Graphing Linear Equations

  3. Lesson 3

    Lesson 3: Using Slope in Problems

  4. Lesson 4

    Lesson 4: Find the Equation

  5. Lesson 5

    Lesson 5: Slope and Intercepts

  6. Lesson 6

    Lesson 6: Comparing Lines