Math Core

Measuring is where geometry meets numbers. Once you can give a segment a length, you can compare segments, split them into pieces and write equations about them, which is how most geometry problems end up being solved.

Length on a number line

Imagine laying a ruler along a line so that every point matches exactly one real number, called its coordinate. This idea is the Ruler Postulate. The distance between two points is then the absolute value of the difference of their coordinates.

Point A has coordinate -3 and point B has coordinate 5.

If $A$ is at $-3$ and $B$ is at $5$ , then

$AB = |5 - (-3)| = |8| = 8.$

The absolute value makes the order irrelevant: $|-3 - 5| = |-8| = 8$ as well. Distance is never negative.

Notice the notation. $\overline{AB}$ (with a bar) is the segment, a set of points. $AB$ (no bar) is its length, a number. You write $AB = 8$ , never $\overline{AB} = 8$ .

Congruent segments

Definition

Congruent segments

Two segments are congruent if they have the same length. You write $\overline{AB} \cong \overline{CD}$ , which means exactly the same thing as $AB = CD$ .

The symbol $\cong$ is for figures and the symbol $=$ is for numbers. In diagrams, matching tick marks on two segments show that they are congruent.

The Segment Addition Postulate

If you walk from $A$ to $C$ and pass through $B$ on the way, the total distance is the sum of the two legs of the trip.

B is between A and C, so AB + BC = AC: 4 + 6 = 10.

Segment Addition Postulate

If $B$ is between $A$ and $C$ (on the same line), then

$AB + BC = AC.$

It also works the other way: if $AB + BC = AC$ for three collinear points, then $B$ is between $A$ and $C$ .

This one fact turns a lot of geometry questions into algebra. You'll write an equation, solve for the variable, and then substitute back to find lengths.

Worked example: Segment addition with algebra

$B$ is between $A$ and $C$ . $AB = 2x + 3$ , $BC = 4x - 1$ and $AC = 26$ . Find $x$ , $AB$ and $BC$ .

By the Segment Addition Postulate, $AB + BC = AC$ :

\begin{aligned} (2x + 3) + (4x - 1) &= 26 \\ 6x + 2 &= 26 \\ x &= 4 \end{aligned}

Then $AB = 2(4) + 3 = 11$ and $BC = 4(4) - 1 = 15$ . Check: $11 + 15 = 26$ .

Common mistake

Finding $x$ is usually not the end of the problem. Read the question again: if it asks for a length, substitute $x$ back into the expression. And always check that every length comes out positive. A negative length means an error.

Worked example: Congruent segments

$\overline{PQ} \cong \overline{RS}$ , with $PQ = 3x - 5$ and $RS = x + 9$ . Find $PQ$ .

Congruent segments have equal lengths, so $3x - 5 = x + 9$ . Then $2x = 14$ and $x = 7$ . So $PQ = 3(7) - 5 = 16$ . Check: $RS = 7 + 9 = 16$ .

Distance in the coordinate plane

When two points aren't on the same horizontal or vertical line, you can't just subtract one coordinate. Instead, draw a right triangle. The horizontal leg is the change in $x$ , the vertical leg is the change in $y$ , and the segment you want is the hypotenuse.

The legs are 7 − 1 = 6 and 10 − 2 = 8, so AB is the hypotenuse of a 6-8-10 right triangle.Open in grapher →

By the Pythagorean theorem, $AB^2 = 6^2 + 8^2 = 100$ , so $AB = 10$ . Doing this with letters instead of numbers gives a formula that works every time.

The Distance Formula

The distance between $(x_1, y_1)$ and $(x_2, y_2)$ is

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

Because the differences are squared, it doesn't matter which point you call the first one, and negative differences cause no trouble.

Worked example: Using the distance formula

Find the distance between $(-2, 5)$ and $(3, -7)$ .

The change in $x$ is $3 - (-2) = 5$ and the change in $y$ is $-7 - 5 = -12$ . So

$d = \sqrt{5^2 + (-12)^2} = \sqrt{25 + 144} = \sqrt{169} = 13.$

Tip

Watch for the Pythagorean triples $3$ - $4$ - $5$ , $5$ - $12$ - $13$ , $8$ - $15$ - $17$ and their multiples (like $6$ - $8$ - $10$ ). If the two legs match one, you know the distance without taking a square root. When they don't, leave the answer as a simplified radical or round it as asked.