# ACT Math: Reviewing Distance Formula

One of the formulas that you’ll want to be familiar with for the ACT Math section is the Distance Formula. It can be used to find the distance, in coordinate units, between two points in the standard (x, y) coordinate plane.

The formula is:

$d = \sqrt{(y_{2}-y_{1})^{2}+(x_{2}-x_{1})^{2}}$

To see how this formula is used, consider the following problem:

What is the distance in the standard (x, y) coordinate plane between the points (8, 3) and (13, 10)?

To answer this question, simply plug the coordinates of the two points into the formula.

$d = \sqrt{(y_{2}-y_{1})^{2}+(x_{2}-x_{1})^{2}}$
$d = \sqrt{(10-3)^{2}+(13-8)^{2}}$
$d = \sqrt{7^{2}+ 5^{2}}$
$d = \sqrt{49 + 25}$
$d = \sqrt{74}$

It turns out, however, that if you forget this formula you can still get the problem correct. All you have to do is recognize that the Distance Formula is really Pythagorean Theorem “in disguise.” Simply sketch the problem, draw in a right triangle and find the length of the hypotenuse, which is the distance between the two points.

$a^{2} + b^{2} = c^{^{2}}$
$7^{2} + 5^{2} = c^{^{2}}$
$49 + 25 = c^{^{2}}$
$74 = c^{^{2}}$
$\sqrt{74} = c$

You’ll get the same answer using either the Distance Formula or by using Pythagorean Theorem. Use the method that makes the most sense to you.

The Distance Formula can also be used to find the radius of a circle if you know the coordinates of its center and the coordinates of a point on the circle. Consider the following problem:

A circle has its center at (3, 4) in the standard (x, y) coordinate plane. If one point on the circle has coordinates (5, 7), what is the length of the radius of the circle?

This question is the same as asking us to find the distance between (3, 4) and (5, 7). Once again, let’s plug into the formula:

$d = \sqrt{(y_{2}-y_{1})^{2}+(x_{2}-x_{1})^{2}}$
$d = \sqrt{(7-4)^{2}+(5-3)^{2}}$
$d = \sqrt{3^{2}+ 2^{2}}$
$d = \sqrt{9 + 4}$
$d = \sqrt{13}$

Alternatively, you could have sketched the problem and used Pythagorean Theorem to get the same answer.

The Distance Formula can also be used to find the length of the diameter of a circle is you know the coordinates of the two endpoints of the diameter.

Here are a few problems that you can practice with. The answers appear at the bottom of this post.

1)  Find the distance in the standard (x, y) coordinate plane between the points (2, 7) and (8, 12).
2)  Find the distance in the standard (x, y) coordinate plane between the points (-4, 9) and (1, 7).
3)  A circle has its center at (2, 9) in the standard (x, y) coordinate plane. If one point on the circle is (-1, 5), what is the length of the radius of the circle?
4)  The endpoints of a diameter of a circle have coordinates, in the standard (x, y) coordinate plane, of (7, 5) and (4, 13). What is the length of the diameter?

If you have questions about these problems or anything else to do with the ACT, leave a comment below or send me an email at info@cardinalec.com.

Solutions:

1)  $\sqrt{61}$
2)  $\sqrt{29}$
3)  $r = 5$
4)  $d = \sqrt{73}$