13.1 Distance and midpoint formulas

The distance and midpoint formulas are the foundational tools for working with line segments in a coordinate plane. They let you find the length between any two points and locate the exact center of a segment. Both formulas come up constantly in coordinate geometry proofs and problems, so getting comfortable with them now pays off throughout the rest of the course.

The distance formula is rooted in the Pythagorean theorem, and the midpoint formula is just an averaging technique applied to coordinates. You'll use them for everything from verifying that a quadrilateral is a rectangle to solving applied problems involving maps and scale.

Distance and Midpoint Formulas

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Distance Formula for Line Segments

The distance formula calculates the length of a line segment between two points in a coordinate plane. It works by treating the segment as the hypotenuse of a right triangle, then applying the Pythagorean theorem ($a^2 + b^2 = c^2$).

For points $(x_1, y_1)$ and $(x_2, y_2)$, the distance $d$ is:

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

Here's why this works: the horizontal distance between the points is $(x_2 - x_1)$ and the vertical distance is $(y_2 - y_1)$. Those form the two legs of a right triangle, and the segment you care about is the hypotenuse. Squaring the differences, adding them, and taking the square root gives you that hypotenuse length.

Steps to find the distance:

1. Identify the coordinates of both endpoints.
2. Subtract the $x$-values to get the horizontal difference, and the $y$-values to get the vertical difference.
3. Square each difference, then add the two squares together.
4. Take the square root of that sum.

Note that it doesn't matter which point you call $(x_1, y_1)$ and which you call $(x_2, y_2)$. Squaring eliminates any negative signs, so the order won't change your answer.

Example: Find the distance between $(3, 4)$ and $(7, 1)$.

• $d = \sqrt{(7 - 3)^2 + (1 - 4)^2} = \sqrt{4^2 + (-3)^2} = \sqrt{16 + 9} = \sqrt{25} = 5$

This is a clean answer because 3-4-5 is a Pythagorean triple. Most problems won't simplify this neatly, so be ready to leave answers in simplified radical form or round to a specified decimal place.

Midpoint Formula for Coordinates

The midpoint formula finds the point that divides a line segment into two equal parts. It does this by averaging: you take the average of the two $x$-coordinates and the average of the two $y$-coordinates.

For endpoints $(x_1, y_1)$ and $(x_2, y_2)$, the midpoint $M$ is:

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

Think of it this way: the midpoint's $x$-coordinate is exactly halfway between the two $x$-values, and the same goes for $y$. That's all an average does here.

Steps to find the midpoint:

1. Identify the coordinates of both endpoints.
2. Add the two $x$-coordinates and divide by 2.
3. Add the two $y$-coordinates and divide by 2.
4. Write the result as an ordered pair $(x, y)$.

Example: Find the midpoint of the segment with endpoints $(-2, 5)$ and $(6, 9)$.

• $x$-coordinate: $\frac{-2 + 6}{2} = \frac{4}{2} = 2$
• $y$-coordinate: $\frac{5 + 9}{2} = \frac{14}{2} = 7$
• Midpoint: $(2, 7)$

A useful check: the midpoint's coordinates should each fall between the corresponding endpoint values. Here, 2 is between $-2$ and 6, and 7 is between 5 and 9, so the answer makes sense.

Applications of Distance and Midpoint

Knowing which formula to use matters just as much as knowing how to use it. A good rule of thumb:

• Use the distance formula when a problem asks for a length, how far apart two things are, or whether two segments are congruent.
• Use the midpoint formula when a problem asks for a center point, a bisector location, or the point equidistant from two endpoints.

In honors geometry, you'll also combine these formulas. For instance, you might find a midpoint first, then use the distance formula to verify that it's equidistant from both endpoints.

Example with scale: Two cities are located at $(120, 250)$ and $(400, 150)$ on a map where 1 unit = 10 miles. Find the distance between them.

1. $d = \sqrt{(400 - 120)^2 + (150 - 250)^2} = \sqrt{280^2 + (-100)^2}$

2. $= \sqrt{78{,}400 + 10{,}000} = \sqrt{88{,}400} \approx 297.3$ units

3. Apply the scale: $297.3 \times 10 \approx 2{,}973$ miles

Always remember to convert your answer using the given scale. The raw number from the formula is in map units, not real-world units, until you multiply.