The Distance Formula and the Pythagorean Theorem

The Pythagorean Theorem: More Than Just Triangles

Before we place anything on a coordinate plane, let's start with the core principle. You probably remember the Pythagorean theorem from geometry, which describes the relationship between the sides of a right-angled triangle.

Algebraically, it's stated as:
$a^2+b^2=c^2$
where $a$ and $b$ are the lengths of the two legs (the sides that form the right angle), and $c$ is the length of the hypotenuse (the side opposite the right angle).

Your goal is to understand the "why," so let's not just state the theorem but also briefly look at a proof. This helps build the intuition that mathematical formulas have logical origins.

From Triangles to Coordinates

Now, how does $a^2+b^2=c^2$ help us find the distance between two points on a graph? Let's visualize it.

Imagine any two points on the coordinate plane, let's call them $P_1$ with coordinates $(x_1,y_1)$ and $P_2$ with coordinates $(x_2,y_2)$. The straight-line distance between them is the length we want to find.

We can draw a horizontal line through one point and a vertical line through the other. Where they intersect, they form a right angle. By connecting our original two points, we create a right-angled triangle, with the distance between the points serving as the hypotenuse.

The lengths of the legs of this triangle, a and b, are simply the differences in the coordinates:

• The length of the horizontal leg (a) is the difference between the x-coordinates: $|x_2-x_1|$.
• The length of the vertical leg (b) is the difference between the y-coordinates: $|y_2-y_1|$.

We use absolute values because length must be a positive quantity.

Now, we can substitute these expressions back into the Pythagorean theorem:
$(|x_2-x_1|{)}^2+(|y_2-y_1|{)}^2=d^2$
where $d$ is the distance (our hypotenuse).

Since squaring any number (positive or negative) results in a positive number, the absolute value bars become redundant. For example, $|-5{|}^2=5^2=25$ and $(-5{)}^2=25$. So, we can simplify the equation to:
$(x_2-x_1{)}^2+(y_2-y_1{)}^2=d^2$
To solve for the distance $d$, we take the square root of both sides. Since distance can't be negative, we only consider the positive root. This gives us the final distance formula.

The Distance Formula

The process we just walked through leads directly to the distance formula. For a more dynamic explanation of this derivation, the following video is quite helpful.

The result of this derivation is the general formula for the distance $d$ between any two points $(x_1,y_1)$ and $(x_2,y_2)$:

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

This formula is a cornerstone of coordinate geometry.

To see the derivation laid out one more time in a clear, graphical format, the following resource is excellent. It neatly presents the entire logical flow from start to finish.

As a final note on terminology, you may sometimes see the differences in coordinates represented by the Greek letter delta ($\Delta$). For instance, $\Delta x=x_2-x_1$ and $\Delta y=y_2-y_1$. Using this notation, the distance formula can be written compactly as $d=\sqrt{(\Delta x{)}^2+(\Delta y{)}^2}$. This is just a different way of writing the same thing.

Let's quickly apply this to find the distance between the points $(2,1)$ and $(6,4)$.

1. Identify your coordinates: $x_1=2,y_1=1$ and $x_2=6,y_2=4$.
2. Calculate the differences:
   • $\Delta x=x_2-x_1=6-2=4$
   • $\Delta y=y_2-y_1=4-1=3$
3. Plug them into the formula:
   $d=\sqrt{4^2+3^2}$
4. Solve:
   $d=\sqrt{16+9}=\sqrt{25}=5$
   Notice that the legs of our triangle have lengths 3 and 4, and the hypotenuse is 5. This is a classic "3-4-5" right triangle, which reinforces the direct connection between the distance formula and the Pythagorean theorem.

Conclusion

In this lesson, we established a critical foundation for your study of coordinate geometry. We didn't just learn a formula; we derived it from first principles.

Here are the key takeaways:

• The distance formula is not an arbitrary rule but a direct application of the Pythagorean theorem to the coordinate plane.
• The distance between two points can be visualized as the hypotenuse of a right triangle.
• The legs of this triangle are the horizontal and vertical distances between the points, found by subtracting their respective coordinates ($x_2-x_1$ and $y_2-y_1$).

You've successfully taken the first step toward building an intuitive grasp of these concepts. In our next lesson, we will put this formula into practice by applying it to solve problems involving points in different quadrants. Afterward, we will use similar logic to derive and apply another essential tool: the midpoint formula.