Learn on PengiSaxon Algebra 1Chapter 9: Quadratic Functions and Equations

Lesson 86: Calculating the Midpoint and Length of a Segment

In this Grade 9 Saxon Algebra 1 lesson, students learn how to apply the distance formula, derived from the Pythagorean Theorem, to calculate the length of a segment between two coordinate points. Students also use the midpoint formula to find the point that divides a line segment into two equal parts. The lesson includes practical applications such as classifying polygons like rhombuses by comparing side lengths using coordinate geometry.

Section 1

πŸ“˜ Calculating the Midpoint and Length of a Segment

New Concept

The distance dd between two points (x1,y1)(x_1, y_1) and (x2,y2)(x_2, y_2) is

d=(x2βˆ’x1)2+(y2βˆ’y1)2d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

What’s next

Next, you’ll apply this formula to find lengths, classify shapes, and pinpoint the exact center of a line segment using the midpoint formula.

Section 2

The Distance Formula

Property

The distance dd between two points (x1,y1)(x_1, y_1) and (x2,y2)(x_2, y_2) is

d=(x2βˆ’x1)2+(y2βˆ’y1)2d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

Explanation

Think of this formula as the Pythagorean theorem's cool cousin for coordinates! It calculates the direct, straight-line distance between any two points by creating a right triangle from the coordinate differences and finding its hypotenuse. It's a secret map shortcut!

Examples

Find the distance between (3,βˆ’2)(3, -2) and (6,4)(6, 4): d=(6βˆ’3)2+(4βˆ’(βˆ’2))2=32+62=45=35d = \sqrt{(6 - 3)^2 + (4 - (-2))^2} = \sqrt{3^2 + 6^2} = \sqrt{45} = 3\sqrt{5}.
Find the distance between (βˆ’1,7)(-1, 7) and (4,βˆ’5)(4, -5): d=(4βˆ’(βˆ’1))2+(βˆ’5βˆ’7)2=52+(βˆ’12)2=169=13d = \sqrt{(4 - (-1))^2 + (-5 - 7)^2} = \sqrt{5^2 + (-12)^2} = \sqrt{169} = 13.

Section 3

Example Card: Finding the Distance Between Two Points

Let's find the exact 'ruler distance' between two points on a map using only their coordinates. This example demonstrates how to use the distance formula, a key idea from this lesson.

Example Problem

Find the distance between (2,βˆ’4)(2, -4) and (7,8)(7, 8).

Section 4

The Midpoint Formula

Property

The midpoint MM of the line segment with endpoints (x1,y1)(x_1, y_1) and (x2,y2)(x_2, y_2) is

M=(x1+x22,y1+y22)M = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)

Explanation

Finding the exact middle of a line is as easy as finding an average, just like the lesson hints! This formula averages the x-values to get the new x-coordinate and averages the y-values for the new y-coordinate, landing you perfectly in the center.

Examples

Find the midpoint of a segment with endpoints (3,5)(3, 5) and (7,βˆ’2)(7, -2): M=(3+72,5+(βˆ’2)2)=(5,32)M = \left(\frac{3 + 7}{2}, \frac{5 + (-2)}{2}\right) = \left(5, \frac{3}{2}\right).
Find the midpoint between (βˆ’2,3)(-2, 3) and (4,7)(4, 7): M=(βˆ’2+42,3+72)=(22,102)=(1,5)M = \left(\frac{-2 + 4}{2}, \frac{3 + 7}{2}\right) = \left(\frac{2}{2}, \frac{10}{2}\right) = (1, 5).

Section 5

Example Card: Finding the Midpoint of a Segment

Finding the dead center of a line is easyβ€”it’s just the average of the endpoints. This example illustrates the midpoint formula, another key idea from this lesson.

Example Problem

Find the midpoint of the line segment with endpoints (2,9)(2, 9) and (8,βˆ’4)(8, -4).

Book overview

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

Continue this chapter

Chapter 9: Quadratic Functions and Equations

  1. Lesson 1

    Lesson 81: Solving Inequalities with Variables on Both Sides

  2. Lesson 2

    Lesson 82: Solving Multi-Step Compound Inequalities

  3. Lesson 3

    Lesson 83: Factoring Special Products

  4. Lesson 4

    Lesson 84: Identifying Quadratic Functions

  5. Lesson 5

    Lesson 85: Solving Problems Using the Pythagorean Theorem

  6. Lesson 6Current

    Lesson 86: Calculating the Midpoint and Length of a Segment

  7. Lesson 7

    Lesson 87: Factoring Polynomials by Grouping

  8. Lesson 8

    Lesson 88: Multiplying and Dividing Rational Expressions

  9. Lesson 9

    Lesson 89: Identifying Characteristics of Quadratic Functions

  10. Lesson 10

    Lesson 90: Adding and Subtracting Rational Expressions

Lesson overview

Expand to review the lesson summary and core properties.

Expand

Section 1

πŸ“˜ Calculating the Midpoint and Length of a Segment

New Concept

The distance dd between two points (x1,y1)(x_1, y_1) and (x2,y2)(x_2, y_2) is

d=(x2βˆ’x1)2+(y2βˆ’y1)2d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

What’s next

Next, you’ll apply this formula to find lengths, classify shapes, and pinpoint the exact center of a line segment using the midpoint formula.

Section 2

The Distance Formula

Property

The distance dd between two points (x1,y1)(x_1, y_1) and (x2,y2)(x_2, y_2) is

d=(x2βˆ’x1)2+(y2βˆ’y1)2d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

Explanation

Think of this formula as the Pythagorean theorem's cool cousin for coordinates! It calculates the direct, straight-line distance between any two points by creating a right triangle from the coordinate differences and finding its hypotenuse. It's a secret map shortcut!

Examples

Find the distance between (3,βˆ’2)(3, -2) and (6,4)(6, 4): d=(6βˆ’3)2+(4βˆ’(βˆ’2))2=32+62=45=35d = \sqrt{(6 - 3)^2 + (4 - (-2))^2} = \sqrt{3^2 + 6^2} = \sqrt{45} = 3\sqrt{5}.
Find the distance between (βˆ’1,7)(-1, 7) and (4,βˆ’5)(4, -5): d=(4βˆ’(βˆ’1))2+(βˆ’5βˆ’7)2=52+(βˆ’12)2=169=13d = \sqrt{(4 - (-1))^2 + (-5 - 7)^2} = \sqrt{5^2 + (-12)^2} = \sqrt{169} = 13.

Section 3

Example Card: Finding the Distance Between Two Points

Let's find the exact 'ruler distance' between two points on a map using only their coordinates. This example demonstrates how to use the distance formula, a key idea from this lesson.

Example Problem

Find the distance between (2,βˆ’4)(2, -4) and (7,8)(7, 8).

Section 4

The Midpoint Formula

Property

The midpoint MM of the line segment with endpoints (x1,y1)(x_1, y_1) and (x2,y2)(x_2, y_2) is

M=(x1+x22,y1+y22)M = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)

Explanation

Finding the exact middle of a line is as easy as finding an average, just like the lesson hints! This formula averages the x-values to get the new x-coordinate and averages the y-values for the new y-coordinate, landing you perfectly in the center.

Examples

Find the midpoint of a segment with endpoints (3,5)(3, 5) and (7,βˆ’2)(7, -2): M=(3+72,5+(βˆ’2)2)=(5,32)M = \left(\frac{3 + 7}{2}, \frac{5 + (-2)}{2}\right) = \left(5, \frac{3}{2}\right).
Find the midpoint between (βˆ’2,3)(-2, 3) and (4,7)(4, 7): M=(βˆ’2+42,3+72)=(22,102)=(1,5)M = \left(\frac{-2 + 4}{2}, \frac{3 + 7}{2}\right) = \left(\frac{2}{2}, \frac{10}{2}\right) = (1, 5).

Section 5

Example Card: Finding the Midpoint of a Segment

Finding the dead center of a line is easyβ€”it’s just the average of the endpoints. This example illustrates the midpoint formula, another key idea from this lesson.

Example Problem

Find the midpoint of the line segment with endpoints (2,9)(2, 9) and (8,βˆ’4)(8, -4).

Book overview

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

Continue this chapter

Chapter 9: Quadratic Functions and Equations

  1. Lesson 1

    Lesson 81: Solving Inequalities with Variables on Both Sides

  2. Lesson 2

    Lesson 82: Solving Multi-Step Compound Inequalities

  3. Lesson 3

    Lesson 83: Factoring Special Products

  4. Lesson 4

    Lesson 84: Identifying Quadratic Functions

  5. Lesson 5

    Lesson 85: Solving Problems Using the Pythagorean Theorem

  6. Lesson 6Current

    Lesson 86: Calculating the Midpoint and Length of a Segment

  7. Lesson 7

    Lesson 87: Factoring Polynomials by Grouping

  8. Lesson 8

    Lesson 88: Multiplying and Dividing Rational Expressions

  9. Lesson 9

    Lesson 89: Identifying Characteristics of Quadratic Functions

  10. Lesson 10

    Lesson 90: Adding and Subtracting Rational Expressions