Math Core

How far apart are two points on a map or a coordinate grid? If they line up horizontally or vertically, you can just count. If they don't, the grid gives you a right triangle for free, and the Pythagorean theorem does the rest.

Horizontal and vertical distances

When two points have the same $y$ -coordinate, they lie on a horizontal line. The distance between them is the difference of their $x$ -coordinates. When they have the same $x$ -coordinate, subtract the $y$ -coordinates.

$(1, 3)$ and $(6, 3)$ : distance $6 - 1 = 5$ .
$(2, -4)$ and $(2, 5)$ : distance $5 - (-4) = 9$ .

Distance is never negative. If you subtract in the "wrong" order, just drop the negative sign: $|1 - 6| = 5$ .

Slanted distances

Now take the points $(1, 2)$ and $(4, 6)$ . The segment between them is slanted, so you can't count squares along it. Instead, draw a horizontal leg and a vertical leg to make a right triangle.

The segment from (1, 2) to (4, 6) is the hypotenuse of a right triangle with legs 3 and 4.Open in grapher →

The horizontal leg goes from $x = 1$ to $x = 4$ , so it is $4 - 1 = 3$ units long.
The vertical leg goes from $y = 2$ to $y = 6$ , so it is $6 - 2 = 4$ units long.
The distance is the hypotenuse: $d = \sqrt{3^2 + 4^2} = \sqrt{25} = 5$ .

Distance between two points

To find the distance between two points:

Find the horizontal change (the difference of the $x$ -coordinates).
Find the vertical change (the difference of the $y$ -coordinates).
Use them as the legs of a right triangle: $d = \sqrt{(\text{horizontal change})^2 + (\text{vertical change})^2}$ .

Worked example: Points in different quadrants

Find the distance between $(-3, 4)$ and $(2, -2)$ , exactly and to the nearest tenth.

The legs meet at the corner (2, 4).Open in grapher →

Horizontal change: from $x = -3$ to $x = 2$ is $2 - (-3) = 5$ units.
Vertical change: from $y = 4$ to $y = -2$ is $4 - (-2) = 6$ units.

$d = \sqrt{5^2 + 6^2} = \sqrt{25 + 36} = \sqrt{61} \approx 7.8$

Common mistake

Be careful when a coordinate is negative. From $x = -3$ to $x = 2$ is $5$ units, not $2 - 3 = -1$ . Subtracting a negative means adding: $2 - (-3) = 2 + 3 = 5$ . You can also count the squares on the grid to check.

The distance formula

If you write the steps above with letters, you get a formula. For points $(x_1, y_1)$ and $(x_2, y_2)$ ,

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

This is just the Pythagorean theorem in disguise: $x_2 - x_1$ is the horizontal leg and $y_2 - y_1$ is the vertical leg. Because each difference is squared, it doesn't matter which point you call the first one. A negative difference becomes positive when you square it.

Worked example: Perimeter of a triangle

A triangle has vertices $A(-2, -1)$ , $B(4, -1)$ and $C(1, 3)$ . Find its perimeter.

Find each side.

$AB$ : the points share $y = -1$ , so $AB = 4 - (-2) = 6$ .
$AC$ : horizontal change $1 - (-2) = 3$ , vertical change $3 - (-1) = 4$ . So $AC = \sqrt{3^2 + 4^2} = 5$ .
$BC$ : horizontal change $4 - 1 = 3$ , vertical change $3 - (-1) = 4$ . So $BC = \sqrt{3^2 + 4^2} = 5$ .

The perimeter is $6 + 5 + 5 = 16$ units. (Two sides are equal, so the triangle is isosceles.)

Tip

Sketch the points before you compute, even roughly. A sketch shows you which way the legs go and gives you a quick estimate to check your answer against.