The distance between two points can be calculated by creating a right triangle with the line segment as the hypotenuse and then using the Pythagorean theorem.

\begin{aligned} d^2&= a^2+b^2 \\ d& = \sqrt{a^2+b^2} \end{aligned}

We can also use the formula which comes from the Pythagorean theorem.

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

\bm{d}

The distance between two points

\bm{\left(x_1,y_1\right)}

The coordinates of the first point

\bm{\left(x_2,y_2\right)}

The coordinates of the second point

Worked examples

Example 1

The points A \left(-3,-2\right), B \left(-3,-4\right), and C \left(1,-4\right) are the vertices of a right triangle as shown on the coordinate plane.

-4

-3

-2

-1

-4

-3

-2

-1

Find AB.

Approach

Since this is a vertical line segment, we can count the squares to calculate this length or find the absolute value of the difference in the y-coordinates.

Solution

\displaystyle AB	\displaystyle =	\displaystyle \left\vert -2--4 \right \vert
	\displaystyle =	\displaystyle 2

Find BC.

Approach

Since this is a horizontal line segment, we can count the squares to calculate this length or find the absolute value of the difference in the x-coordinates.

Solution

\displaystyle BC	\displaystyle =	\displaystyle \left\vert -3-1 \right \vert
	\displaystyle =	\displaystyle 4

Find AC using the Pythagorean theorem, rounding to three decimal places.

Approach

AC is our hypotenuse and we have the lengths of other two sides AB=2 and BC=4.

Solution

\displaystyle AC^2	\displaystyle =	\displaystyle AB^2+BC^2	Pythagorean theorem
\displaystyle AC^2	\displaystyle =	\displaystyle 2^2+4^2	Substitute
\displaystyle AC^2	\displaystyle =	\displaystyle 4+16	Evaluate the squares
\displaystyle AC^2	\displaystyle =	\displaystyle 20	Evaluate the addition
\displaystyle AC	\displaystyle =	\displaystyle \sqrt{20}	Take the square root of both sides
\displaystyle AC	\displaystyle =	\displaystyle 4.472	Round using a calculator

Example 2

Find the distance between A \left(-1,9\right) and B \left(-4,1\right). Leave your answer in exact (radical) form.

Approach

We can either plot \overline{AB} on the coordinate plane and use the Pythagorean theorem, or we can use the distance formula. Using the formula will likely be more efficient.

Solution

We will use A \left(-1,9\right) as \left(x_1,y_1\right) and B \left(-4,1\right) as \left(x_2,y_2\right).

\displaystyle d	\displaystyle =	\displaystyle \sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}	Distance formula
	\displaystyle =	\displaystyle \sqrt{\left(-4--1\right)^2+\left(1-9\right)^2}	Substitute
	\displaystyle =	\displaystyle \sqrt{\left(-3\right)^2+\left(-8\right)^2}	Simplify the addition in the parentheses
	\displaystyle =	\displaystyle \sqrt{9+64}	Evaluate the squares
	\displaystyle =	\displaystyle \sqrt{73}	Evaluate the addition

The distance between A and B is \sqrt{73} units.

Reflection

There are no perfect squares as factors of 73, so there is no further simplification to do.

A good final check is that our answer is positive as it is a length, so it must be a positive value.

We could also have used A \left(-1,9\right) as \left(x_2,y_2\right) and B \left(-4,1\right) as \left(x_1,y_1\right) and gotten the same answer.

Outcomes

G.N.Q.A.1

Use units as a way to understand real-world problems.*

G.N.Q.A.1.A

Use appropriate quantities in formulas, converting units as necessary.

G.GPE.A.3

Understand the relationship between the Pythagorean Theorem and the distance formula and use an efficient method to solve problems on the coordinate plane.

G.MP1

Make sense of problems and persevere in solving them.

G.MP2

Reason abstractly and quantitatively.

G.MP3

Construct viable arguments and critique the reasoning of others.

G.MP4

Model with mathematics.

G.MP5

Use appropriate tools strategically.

G.MP6

Attend to precision.

G.MP7

Look for and make use of structure.

G.MP8

Look for and express regularity in repeated reasoning.

9.01 Distance and the Pythagorean theorem

Concept summary

Worked examples

Example 1

Approach

Solution

Approach

Solution

Approach

Solution

Example 2

Approach

Solution

Reflection

Outcomes

G.N.Q.A.1

G.N.Q.A.1.A

G.GPE.A.3

G.MP1

G.MP2

G.MP3

G.MP4

G.MP5

G.MP6

G.MP7

G.MP8

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