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

MA.912.T.1.2

Solve mathematical and real-world problems involving right triangles using trigonometric ratios and the Pythagorean Theorem.

10.01 Distance and the Pythagorean theorem

Concept summary

Worked examples

Example 1

Approach

Solution

Approach

Solution

Approach

Solution

Example 2

Approach

Solution

Reflection

Outcomes

MA.912.T.1.2

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