If $A(x_1,y_1)$ and $B(x_2,y_2)$ are two points in a coordinate plane, the distance between them, $AB,$ can be found.

The horizontal and vertical distance between the points can be thought of as $\Delta x$ and $\Delta y,$ respectively.

As can be seen in the diagram above, $\overline{AB},$ which gives the distance between points $A$ and $B,$ is the hypotenuse of the right triangle whose legs are $\Delta x$ and $\Delta y.$ An equation can be created for $AB$ using the Pythagorean Theorem, $a^2+b^2=c^2.$ $$(\Delta x{)}^2+(\Delta y{)}^2=(AB{)}^2$$ $\Delta x$ and $\Delta y$ are the differences between the coordinates of the points. They can be written as $\Delta x=\left|x_2-x_1\right|$ and $\Delta y=\left|y_2-y_1\right|.$ Here, the absolute values are included to ensure the distances are positive. Rearranging the equation and substituting these values gives the following.

Because squares are always positive, the absolute value bars can be removed. Then, $AB$ can be isolated.

Since $AB$ is a distance, it is always positive. Therefore, the negative solution can be omitted. $$AB=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}$$