Learn on PengiBig Ideas Math, Advanced 2Chapter 7: Real Numbers and the Pythagorean Theorem

Section 7.5: Using the Pythagorean Theorem

Property The distance $d$ between points $P 1(x 1, y 1)$ and $P 2(x 2, y 2)$ is $$d = \sqrt{(x 2 x 1)^2 + (y 2 y 1)^2}$$ This formula is derived from the Pythagorean theorem, where the distance $d$ is the hypotenuse of a right triangle whose legs have lengths $|x 2 x 1|$ and $|y 2 y 1|$.

Section 1

Distance Formula with Pythagorean Derivation

Property

The distance $d$ between points $P_1(x_1, y_1)$ and $P_2(x_2, y_2)$ is

d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

This formula is derived from the Pythagorean theorem, where the distance

d

is the hypotenuse of a right triangle whose legs have lengths

|x_2 - x_1|

and

|y_2 - y_1|

Examples

To find the distance between $(1, 3)$ and $(5, 6)$ , we calculate $d = \sqrt{(5 - 1)^2 + (6 - 3)^2} = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5$ .
The distance between $(-2, 7)$ and $(3, -5)$ is $d = \sqrt{(3 - (-2))^2 + (-5 - 7)^2} = \sqrt{5^2 + (-12)^2} = \sqrt{25 + 144} = \sqrt{169} = 13$ .
The distance between $(4, -1)$ and $(-5, -1)$ is $d = \sqrt{(-5 - 4)^2 + (-1 - (-1))^2} = \sqrt{(-9)^2 + 0^2} = \sqrt{81} = 9$ .

Explanation

Think of this as the Pythagorean theorem on a coordinate plane. The horizontal change (run) and vertical change (rise) between two points form the legs of a right triangle. The distance formula simply calculates the length of the hypotenuse.

Section 2

Distance Formula

Property

The distance between two points is the length of the segment joining them. The distance $d$ between points $P_1(x_1, y_1)$ and $P_2(x_2, y_2)$ is

d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

Examples

The distance between $(2, 3)$ and $(5, 7)$ is $d = \sqrt{(5 - 2)^2 + (7 - 3)^2} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5$ .

Book overview

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Chapter 7: Real Numbers and the Pythagorean Theorem

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Lesson 1
Section 7.1: Finding Square Roots
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Lesson 2
Section 7.2: Finding Cube Roots
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Lesson 3
Section 7.3: The Pythagorean Theorem
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Lesson 4
Section 7.4: Approximating Square Roots
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Lesson 5Current
Section 7.5: Using the Pythagorean Theorem
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Section 1

Distance Formula with Pythagorean Derivation

Property

The distance $d$ between points $P_1(x_1, y_1)$ and $P_2(x_2, y_2)$ is

d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

This formula is derived from the Pythagorean theorem, where the distance

d

is the hypotenuse of a right triangle whose legs have lengths

|x_2 - x_1|

and

|y_2 - y_1|

Examples

To find the distance between $(1, 3)$ and $(5, 6)$ , we calculate $d = \sqrt{(5 - 1)^2 + (6 - 3)^2} = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5$ .
The distance between $(-2, 7)$ and $(3, -5)$ is $d = \sqrt{(3 - (-2))^2 + (-5 - 7)^2} = \sqrt{5^2 + (-12)^2} = \sqrt{25 + 144} = \sqrt{169} = 13$ .
The distance between $(4, -1)$ and $(-5, -1)$ is $d = \sqrt{(-5 - 4)^2 + (-1 - (-1))^2} = \sqrt{(-9)^2 + 0^2} = \sqrt{81} = 9$ .

Explanation

Section 2

Distance Formula

Property

The distance between two points is the length of the segment joining them. The distance $d$ between points $P_1(x_1, y_1)$ and $P_2(x_2, y_2)$ is

d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

Examples

The distance between $(2, 3)$ and $(5, 7)$ is $d = \sqrt{(5 - 2)^2 + (7 - 3)^2} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5$ .

Book overview

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

Continue this chapter

Chapter 7: Real Numbers and the Pythagorean Theorem

Back to Big Ideas Math, Advanced 2

Lesson 1
Section 7.1: Finding Square Roots
Learn Practice
Lesson 2
Section 7.2: Finding Cube Roots
Learn Practice
Lesson 3
Section 7.3: The Pythagorean Theorem
Learn Practice
Lesson 4
Section 7.4: Approximating Square Roots
Learn Practice
Lesson 5Current
Section 7.5: Using the Pythagorean Theorem
Learn Practice

Overview

Lesson overview

Review the lesson summary and core properties without leaving the current page.

Overview

Section 1

Distance Formula with Pythagorean Derivation

Property

The distance $d$ between points $P_1(x_1, y_1)$ and $P_2(x_2, y_2)$ is

d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

This formula is derived from the Pythagorean theorem, where the distance

d

is the hypotenuse of a right triangle whose legs have lengths

|x_2 - x_1|

and

|y_2 - y_1|

Examples

To find the distance between $(1, 3)$ and $(5, 6)$ , we calculate $d = \sqrt{(5 - 1)^2 + (6 - 3)^2} = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5$ .
The distance between $(-2, 7)$ and $(3, -5)$ is $d = \sqrt{(3 - (-2))^2 + (-5 - 7)^2} = \sqrt{5^2 + (-12)^2} = \sqrt{25 + 144} = \sqrt{169} = 13$ .
The distance between $(4, -1)$ and $(-5, -1)$ is $d = \sqrt{(-5 - 4)^2 + (-1 - (-1))^2} = \sqrt{(-9)^2 + 0^2} = \sqrt{81} = 9$ .

Explanation

Section 2

Distance Formula

Property

The distance between two points is the length of the segment joining them. The distance $d$ between points $P_1(x_1, y_1)$ and $P_2(x_2, y_2)$ is

d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

Examples

The distance between $(2, 3)$ and $(5, 7)$ is $d = \sqrt{(5 - 2)^2 + (7 - 3)^2} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5$ .

Book overview

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

Continue this chapter

Chapter 7: Real Numbers and the Pythagorean Theorem

Back to Big Ideas Math, Advanced 2

Lesson 1
Section 7.1: Finding Square Roots
Learn Practice
Lesson 2
Section 7.2: Finding Cube Roots
Learn Practice
Lesson 3
Section 7.3: The Pythagorean Theorem
Learn Practice
Lesson 4
Section 7.4: Approximating Square Roots
Learn Practice
Lesson 5Current
Section 7.5: Using the Pythagorean Theorem
Learn Practice

Section 7.5: Using the Pythagorean Theorem

Property

Examples

Explanation

Property

Examples

Chapter 7: Real Numbers and the Pythagorean Theorem

Section 7.1: Finding Square Roots

Section 7.2: Finding Cube Roots

Section 7.3: The Pythagorean Theorem

Section 7.4: Approximating Square Roots

Section 7.5: Using the Pythagorean Theorem

Lesson overview

Property

Examples

Explanation

Property

Examples

Chapter 7: Real Numbers and the Pythagorean Theorem

Section 7.1: Finding Square Roots

Section 7.2: Finding Cube Roots

Section 7.3: The Pythagorean Theorem

Section 7.4: Approximating Square Roots

Section 7.5: Using the Pythagorean Theorem

Section 7.1: Finding Square Roots

Section 7.2: Finding Cube Roots

Section 7.3: The Pythagorean Theorem

Section 7.4: Approximating Square Roots

Section 7.5: Using the Pythagorean Theorem