Distance on a Number Line and in the Plane | Geometry | LevelKite

How far apart are two points? The distance between two points is the length of the segment that joins them. A distance is never negative, and it equals $0$ only when the two points are the same point.

In this lesson, you find distances exactly from coordinates, the numbers that give a point's position. You start on a number line and then move to the coordinate plane. Every value comes from the stated coordinates, never from how a drawing looks.

Start with what distance means on a number line. The number line below shows Point A at coordinate $-3$ and Point B at coordinate $5$ . You write $AB$ for the distance between them. To get from Point A to Point B, you move $3$ unit steps to reach $0$ , then $5$ more unit steps to reach $5$ . That is $3 + 5 = 8$ unit steps, so $AB = 8$ .

Point A is at

-3

and Point B is at

5

, so

8

unit steps separate them.

Counting steps gets hard when coordinates are fractions, so you need a rule. The Ruler Postulate gives it. A postulate is a statement accepted without proof. This one says that the points on a line can be paired one-to-one with the real numbers, where each point's number is its coordinate. It also says that the distance between two points is the absolute value of the difference of their coordinates. The absolute value of a number is the number itself if it is nonnegative, and its opposite if it is negative, so it is never negative.

Why use absolute value? The two orders of subtraction give opposite numbers. For Points A and B, $5 - (-3) = 8$ and $-3 - 5 = -8$ . The absolute value of each is $8$ , which matches your count. So either order gives the distance. In general, if point $A$ has coordinate $a$ and point $B$ has coordinate $b$ , where $a$ and $b$ are any real numbers, then:

AB = |b - a| = |a - b|

Worked example

Two negative coordinates

On a number line, point $C$ has coordinate $-7$ and point $D$ has coordinate $-2$ . Find $CD$ .

Apply the Ruler Postulate: the distance is the absolute value of the difference of the coordinates. Substitute the coordinates of D and C.
$CD = |-2 - (-7)|$
Subtracting a number is the same as adding its opposite, so subtracting negative 7 means adding 7.
$CD = |-2 + 7| = |5|$
Take the absolute value. The absolute value of a positive number is the number itself.
$CD = 5$
Check with the other order of subtraction. It gives the same distance. The distance is also positive, as it must be for two different points.
$|-7 - (-2)| = |-7 + 2| = |-5| = 5$

Answer

The distance between $C$ and $D$ is $5$ units, so $CD = 5$ .

Worked example

Fraction coordinates

On a number line, point $E$ has coordinate $-\frac{3}{2}$ and point $F$ has coordinate $\frac{5}{4}$ . Find $EF$ exactly.

Apply the Ruler Postulate and substitute the coordinates of F and E.
$EF = \left|\frac{5}{4} - \left(-\frac{3}{2}\right)\right|$
Subtracting a number is the same as adding its opposite.
$EF = \left|\frac{5}{4} + \frac{3}{2}\right|$
Rewrite 3/2 as 6/4 so both fractions have denominator 4, then add the numerators. Keep the fraction so the answer stays exact.
$EF = \left|\frac{5}{4} + \frac{6}{4}\right| = \left|\frac{11}{4}\right|$
Take the absolute value. The result is positive, as a distance between two different points must be.
$EF = \frac{11}{4}$

Answer

The exact distance between $E$ and $F$ is $\frac{11}{4}$ units.

Now move to the coordinate plane. It is built from a horizontal number line, the $x$ -axis, and a vertical number line, the $y$ -axis. The axes are perpendicular, meaning they meet at right angles, and they use the same unit. Each point has an ordered pair of coordinates $(x, y)$ .

Take $P(1, 1)$ and $Q(5, 4)$ . They differ in both coordinates, so neither the $x$ -values nor the $y$ -values alone work as number-line coordinates for segment $PQ$ . Instead, add a third point, $R(5, 1)$ . It shares its $y$ -coordinate with $P$ , so segment $PR$ is horizontal. It shares its $x$ -coordinate with $Q$ , so segment $RQ$ is vertical. Because the axes are perpendicular, every horizontal line is perpendicular to every vertical line. So the angle at $R$ is a right angle, and triangle $PRQ$ is a right triangle, a triangle with a right angle. You know this from the coordinates, not from how the figure looks.

The figure below marks the three points and labels each with its coordinates. It does not draw the segments, so sketch them yourself to see the triangle.

P

is at

(1, 1)

,

Q

is at

(5, 4)

, and the added point

R

is at

(5, 1)

.

Now find the side lengths. Segment $PR$ lies on a horizontal line, where the $x$ -values act as number-line coordinates. By the Ruler Postulate, $PR = |5 - 1| = 4$ . Segment $RQ$ lies on a vertical line, where the $y$ -values act as number-line coordinates, so $RQ = |4 - 1| = 3$ .

The Pythagorean Theorem says: if a triangle is a right triangle with legs (the two sides that form the right angle) of lengths $a$ and $b$ and hypotenuse (the side opposite the right angle) of length $c$ , then $a^2 + b^2 = c^2$ . You use this direction, not the converse, which swaps the if and then parts: you know the angle at $R$ is right, and you conclude a fact about the lengths. Segment $PQ$ is the hypotenuse, so $PQ^2 = 4^2 + 3^2 = 16 + 9 = 25$ .

The equation $PQ^2 = 25$ has two solutions for $PQ$ : the positive root $\sqrt{25} = 5$ and the negative root $-\sqrt{25} = -5$ . A distance cannot be negative, so $PQ = 5$ .

The same reasoning works for any two points that differ in both coordinates.

Given: points $P(x_1, y_1)$ and $Q(x_2, y_2)$ with $x_1 \ne x_2$ and $y_1 \ne y_2$ .

To show: $PQ = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$ .

Let $R$ be the point $(x_2, y_1)$ . Segment $PR$ is horizontal because $P$ and $R$ share the $y$ -coordinate $y_1$ , and segment $RQ$ is vertical because $R$ and $Q$ share the $x$ -coordinate $x_2$ . Because the axes are perpendicular, the angle at $R$ is a right angle. By the Ruler Postulate, $PR = |x_2 - x_1|$ and $RQ = |y_2 - y_1|$ . By the Pythagorean Theorem, $PQ^2 = |x_2 - x_1|^2 + |y_2 - y_1|^2$ . A real number and its absolute value have the same square, so $PQ^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2$ . This sum is positive, so the equation has two solutions for $PQ$ , the positive and the negative root. A distance cannot be negative, so $PQ$ is the positive root, the one the square root symbol names.

Key concept

Distance comes from coordinate differences

On a number line, the distance between two points is the absolute value of the difference of their coordinates, by the Ruler Postulate. In the coordinate plane, the Distance Formula below gives the distance between any two points $P(x_1, y_1)$ and $Q(x_2, y_2)$ . Its radicand, the expression under the root symbol, is a sum of squares, so it is never negative, and the root symbol names the nonnegative root. Use the stated coordinates, never the look of a figure. Keep answers exact, with fractions exact and radicals simplified, and label any decimal as an approximation.

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

Worked example

Negative coordinates and a radical

Find the exact distance between $J(-2, 3)$ and $K(4, -1)$ .

Use the Distance Formula, with J as the first point and K as the second. Either order works, because reversing a difference changes only its sign, and its square stays the same.
$JK = \sqrt{(4 - (-2))^2 + (-1 - 3)^2}$
Subtract inside each set of parentheses. Subtracting negative 2 is the same as adding 2.
$JK = \sqrt{6^2 + (-4)^2}$
Square each difference, then add. A negative number times itself is positive, so the square of negative 4 is positive 16. The radicand, 52, is positive, so its square root is a real number.
$JK = \sqrt{36 + 16} = \sqrt{52}$
Simplify the radical by factoring out the perfect square 4. The product property of radicals applies because 4 and 13 are nonnegative.
$JK = \sqrt{4 \cdot 13} = \sqrt{4} \cdot \sqrt{13} = 2\sqrt{13}$
Check the result. It is positive, as a distance between two different points must be. Also, 13 has no perfect-square factor greater than 1, so the radical is fully simplified.

Answer

The exact distance between $J$ and $K$ is $2\sqrt{13}$ units, which is approximately $7.21$ units.

Worked example

A fraction coordinate in the plane

Find the exact distance between $M\left(\frac{1}{2}, -1\right)$ and $N(3, 2)$ .

Substitute into the Distance Formula, with M first and N second.
$MN = \sqrt{\left(3 - \frac{1}{2}\right)^2 + (2 - (-1))^2}$
Subtract inside each set of parentheses. Write 3 as 6/2 to subtract the fraction. Subtracting negative 1 means adding 1, so the second difference is 3.
$MN = \sqrt{\left(\frac{6}{2} - \frac{1}{2}\right)^2 + 3^2} = \sqrt{\left(\frac{5}{2}\right)^2 + 3^2}$
Square each difference. Then write 9 as 36/4 so you can add the fractions.
$MN = \sqrt{\frac{25}{4} + 9} = \sqrt{\frac{25}{4} + \frac{36}{4}} = \sqrt{\frac{61}{4}}$
Use the quotient property of radicals, which applies because 61 is nonnegative and 4 is positive. The square root of 4 is 2.
$MN = \frac{\sqrt{61}}{\sqrt{4}} = \frac{\sqrt{61}}{2}$
Check: 61 has no perfect-square factor greater than 1, so the radical is simplified. The result is positive, as a distance between two different points must be.

Answer

The exact distance between $M$ and $N$ is $\frac{\sqrt{61}}{2}$ units, which is approximately $3.91$ units.

Distance comes from differences in coordinates. On a number line, take the absolute value of the difference. In the coordinate plane, use the Distance Formula, which combines the horizontal and vertical differences through the Pythagorean Theorem. Keep your answer exact and check that it makes sense as a distance: positive for two different points. Next, you use coordinates to find the midpoint of a segment, and to find an endpoint when you know the midpoint.