Distance Formula

$$\textbf{The Distance Formula}$$

The Distance Formula can be used to find the distance between two points on the coordinate plane.

Let's say you have any two points, $(x_1, y_1)$ and $(x_2, y_2),$ and you want to find the distance, $d,$ between them. Then just use the following formula:

The Distance Formula is a literal equation with several variables representing the locations of the points.

For example, say you're given two points, $(3, 7)$ and $(-2, 15),$ and you're asked to find the distance between them.

Step One: Assign $x_1, y_1, x_2,$ and $y_2$

Let's say that $(x_1,y_1)=(3,7)$ and $(x_2,y_2)=(-2,15).$ Thus, $$\begin{align} x_1&=3\\ y_1&=7\\ x_2&=-2\\ x_2&=15 \end{align}$$

Step Two: Substitute Into the Distance Formula

Now that we have $x_1, y_1, x_2,$ and $y_2,$ we can substitute these values into the Distance Formula. $$\begin{align} d&=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\\ &=\sqrt{(-2-3)^2+(15-7)^2}\\ &=\sqrt{(-5)^2+(8)^2}\\ &=\sqrt{25+64}\\ &=\sqrt{89} \end{align}$$

So, the distance, $d,$ between $(3, 7)$ and $(-2, 15)$ is $\sqrt{89}$ units.

To find the distance between two points $(x_{1},y_{1})$ and $(x_{2},y_{2})$ on the coordinate plane, you can use this formula:

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

For example, the distance between $(0,0)$ and $(3,4)$ is equal to $\sqrt{(3-0)^2+(4-0)^2}$, or $\sqrt{3^2+4^2}$, or $5$.

First, let's plug our values into the distance formula.

$$\sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}\\ \sqrt{(5-1)^2+(8-2)^2}$$

Now, let's calculate our answer.

$$\sqrt{(5-1)^2+(8-2)^2} \\ \sqrt{4^2 + 6^2} \\ \sqrt{16 + 36} \\ \sqrt{52}$$

We need to simplify $\sqrt{52}$.

$$\sqrt{52}\\ \sqrt{4} \cdot \sqrt{13}\\ 2\sqrt{13}$$

The final, simplified answer is $2\sqrt{13}$.

Finding the Distance Between $2$ Points on a Line

Although you might see a few different types of problems asking about the distance between $2$ points, but there's only one formula you need:

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

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The $x$ and $y$ terms in the distance equation are the two coordinate pairs or points on the line, each written in the form $(x, y)$.

So, let's put this formula into use!

$$\sqrt{(4-1)^2+(0-4)^2}$$ $$=\sqrt{(3)^2+(-4)^2}$$ $$=\sqrt{9+16}$$ $$=\sqrt{25}$$ $$=5$$

You should note that it does not matter which coordinate pair you pick as $1$ and which you pick as $2$. You'll get the same answer.

You should also note that if you have a negative number under the square root sign after you've squared both sets of terms, you need to go back and check your work. Remember that all squared values are nonnegative.

$$\sqrt{(0-2)^2+(-3-6)^2}$$ $$=\sqrt{(2)^2+(9)^2}$$ $$=\sqrt{4+81}$$ $$=\sqrt{85}$$

The Distance Formula is based on the Pythagorean Theorem. If you know the Pythagorean Theorem, you already know the Distance Formula! Here is how:

Pick any two different points on a grid and draw a line between them. This line will be the hypotenuse of your new right triangle! The length of this line is also the distance between the points.

I will choose these two points, $A$ and $B$, at $(2, 3)$ and $(5, 7)$:

Next, draw the legs of the triangle by drawing a line straight down from the highest point of your line until it is level with the lowest point of your line. Then, draw a line straight across from the lowest point of your line until it intersects with your vertical line. I will do it here in purple, and label the point of intersection of the legs point $C$:

Notice that the $x$-coordinate of $C$ is equal to the $x$-coordinate of $B$, and the $y$-coordinate of $C$ is equal to the $y$-coordinate of $A$!

From here, to find the distance between the points $A$ and $B$, we use the Pythagorean formula, $$c^2=a^2+b^2.$$ (Be careful not to get the variables mixed up! Recall that $a$ and $b$ are the lengths of the two legs. In our current example, these lengths will be the lengths of the two purple lines.)

Let $a$ be the length of the horizontal purple leg. We can see from either counting the grids or finding the difference of the $x$-coordinates of $C$ and $A$, $5-2$, is the length $3$.

Let $b$ be the length of the vertical purple leg. From finding the difference of the $y$-coordinates of $B$ and $C$, we can see that the length is $7-3 = 4$.

Now to solve for $c$ (our green line), we will first find $c^2$ by using the Pythagorean formula. $$c^2 = 3^2 + 4^2 = 9 + 16 = 25$$ Take the square root to get $c = 5$. And that's the distance! We're finished!

Let's go through our steps again, this time with variables, to arrive at the distance formula.

Say point A is at $(X_1,Y_1)$, and point B is at $(X_2, Y_2)$.

Then the length of the horizontal leg ($a$) will be $(X_2-X_1)$ and the length of the vertical leg ($b$) will be $(Y_2-Y_1)$.

Plug these values into the formula to get $c^2 = (X_2-X_1)^2 + (Y_2-Y_1)^2$.

Now we want to solve for c, the distance, so we square root the whole thing.
$$c = \sqrt{\left(X_2-X_1\right)^2+\left(Y_2-Y_1\right)^2}$$ and that's the Distance Formula!

(Video) Distance Formula

by yaymath

Here's a video by yaymath explaining the distance formula between two points.

$\large{\text{Summary}}$

Plus he does it upside down so that's interesting. The distance formula gives you the distance between any two points. The formula is: $$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$$

(Video) Finding the Distance Between Two Points Using the DIstance Formula

by The Math Sorcerer

This video by The Math Sorcerer shows how to find the distance between two points.

$\large{\text{Summary}}$

To find the distance between $(4,1)$ and $(2,3)$ we use the distance formula. $$d=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$$

Now, we just plug in the points and solve. $$\begin{align} d&=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\\ &=\sqrt{(4-2)^2+(1-3)^2}\\ &=\sqrt{(2)^2+(-2)^2}\\ &=\sqrt{4+4}\\ &=\sqrt{8}\\ &=2\sqrt{2} \end{align}$$

(Video) Distance Formula

by GVSUmath

This video by GVSUmath explains the distance formula for two points.

We know the distance formula is $$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$$

Now, let's plug in the points and solve! $$\begin{align} d&=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\\ &=\sqrt{(8-5)^2+(5-1)^2}\\ &=\sqrt{(3)^2+(4)^2}\\ &=\sqrt{9+16}\\ &=\sqrt{25}\\ &=5 \end{align}$$

Now, we look at some points on a coordinate plane. We are given the points $(0,0)$ and $(4,3)$. Plug those into the equation and check your solution after clicking the continue button.

$$\begin{align} d&=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\\ &=\sqrt{(4-0)^2+(3-0)^2}\\ &=\sqrt{(4)^2+(3)^2}\\ &=\sqrt{16+9}\\ &=\sqrt{25}\\ &=\color{red}{5} \end{align}$$

(Video) Distance and Midpoint

by mathman1024

In this video by mathman1024 he goes over the distance formula and the midpoint formula for two points.

$\large{\text{Summary}}$

The distance formula is derived from the Pythagorean Theorem.

As you can see the distance between the points is the hypotenuse. So we can rewrite the Pythagorean Theorem to solve for $c$. $$c=\sqrt{a^2+b^2}$$ You might also notice that $a$ and $b$ are the difference between the $x$ and $y$ coordinates. So the distance $d$ between two points is: $$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$$ Note: It doesn't matter which point you make $(x_1,y_1)$ and which you make $(x_2,y_2)$ as long as you're consistent when using the formula.

The second formula is the midpoint formula. This is super easy to find. The midpoint is a set of coordinates though so it'll be in the form $(x,y)$. To find it you just take the average of the $x$-coordinates and the average of the $y$-coordinates. $$\text{Midpoint}=\bigg(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\bigg)$$

He goes over a problem to show how these are used. The problem is: Find the distance and midpoint between the points $P(4,-7)$ and $Q(-2,1)$. $$\begin{align} d&=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\\ &=\sqrt{(4-(-2))^2+((-7)-1)^2}\\ &=\sqrt{(6)^2+(-8)^2}\\ &=\sqrt{36+64}\\ &=\sqrt{100}\\ &=10 \end{align}$$ So our distance between the points $P(4,-7)$ and $Q(-2,1)$ is $10$ units. Now let's find the midpoint. $$\begin{align} \text{Midpoint}&=\bigg(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\bigg)\\ &=\bigg(\frac{(4)+(-2)}{2},\frac{(-7)+(1)}{2}\bigg)\\ &=\bigg(\frac{2}{2},\frac{-6}{2}\bigg)\\ &=(1,-3) \end{align}$$