Distance Formula & Midpoint Formula

The Distance Formula

Review: The Pythagorean Formula

In both beginning and intermediate algebra, one of the topics introduced is the pythagorean formula. This formula states that for any right triangle, if c is the lenth of the hypotenuse (the longest side), and a and b are the lengths of the other two sides, then:

LaTeX: c^2=a^2+b^2 $c^2=a^2+b^2$

We use this formula to derive the Distance Formula.

Consider the following scenario. We have two points P = $LaTeX: _{\left(x_1,y_1\right)}$ $_{\left(x_1,y_1\right)}$ and R = $LaTeX: \left(x_2,y_2\right)$ $\left(x_2,y_2\right)$ . We want to find the distance between P and R.

First we start by drawing a right triangle, with the corner with the right angle at Q = $LaTeX: \left(x_2,y_1\right)$ $\left(x_2,y_1\right)$ . We notice that the side with length d, which is opposite the point Q, is the hypotenuse. The lengths of the other two sides are $LaTeX: \mid x_2-x_1\mid$ $\mid x_2-x_1\mid$ and $LaTeX: \mid y_2-y_1\mid$ $\mid y_2-y_1\mid$ . See the figure below:

Graph showing two points with a right triangle drawn by going horizontally from one point and vertically from the other.PNG

Notice that, since we have created a right triangle, we can apply the Pythagorean formula with $LaTeX: a=\mid x_2-x_{1\mid}\mid$ $a=\mid x_2-x_{1\mid}\mid$ and $LaTeX: b=\mid y_2-y_1\mid$ $b=\mid y_2-y_1\mid$ and LaTeX: c=d. $c=d.$ This gives us,

$LaTeX: d^2=\mid x_2-x_1\mid^2+\mid y_2-y_1\mid^2=\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2$ $d^2=\mid x_2-x_1\mid^2+\mid y_2-y_1\mid^2=\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2$

And when we consider that d is a distance so it is always non-negative, and we take the square root of both sides of this equation, we get the distance formula.

Distance Formula

$LaTeX: d=\sqrt[]{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}$ $d=\sqrt[]{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}$

Example:

Find the distance between the points $LaTeX: \left(-2,3\right)$ $\left(-2,3\right)$ and $LaTeX: \left(5,7\right)$ $\left(5,7\right)$ .

Solution:

$LaTeX: d=\sqrt[]{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}=\sqrt[]{\left(-2-1\right)^2+\left(5-7\right)^2}$ $d=\sqrt[]{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}=\sqrt[]{\left(-2-1\right)^2+\left(5-7\right)^2}$

$LaTeX: =\sqrt[]{\left(-3\right)^2+\left(-2\right)^2}=\sqrt[]{9+4}=\sqrt[]{13}$ $=\sqrt[]{\left(-3\right)^2+\left(-2\right)^2}=\sqrt[]{9+4}=\sqrt[]{13}$

In this case, the exact answer would be that the distance between the points is $LaTeX: \sqrt[]{13}$ $\sqrt[]{13}$ units. This answer is an irrational number, but if we want a decimal answer we can approximate the $LaTeX: \sqrt[]{13}$ $\sqrt[]{13}$ by using a calculator, and we get ~3.61.

The Midpoint Formula

The midpoint formula helps us find the midpoint of a line segment connecting point $LaTeX: \left(x_1,y_1\right)$ $\left(x_1,y_1\right)$ to point $LaTeX: \left(x_2,y_2\right)$ $\left(x_2,y_2\right)$ . To do this what we need to do is to find the mean of the two x coordinates and the mean of the two y coordinates.

Midpoint = $LaTeX: \left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)$ $\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)$

Example:

Find the midpoint of a line segment connecting $LaTeX: \left(-3,-7\right)$ $\left(-3,-7\right)$ and $LaTeX: \left(1,4\right)$ $\left(1,4\right)$ .

Solution:

Midpoint = $LaTeX: \left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)=$ $\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)=$ $LaTeX: \left(\frac{-3+1}{2},\:\frac{-7+4}{2}\right)=\left(\frac{-2}{2},\frac{-3}{2}\right)=\left(-1,\frac{-3}{2}\right)$ $\left(\frac{-3+1}{2},\:\frac{-7+4}{2}\right)=\left(\frac{-2}{2},\frac{-3}{2}\right)=\left(-1,\frac{-3}{2}\right)$