8th Grade · 8.G.B.8

Distance between two points in 8th grade

In 8th grade math, finding the distance between two points means working out the length of a slanted line on a coordinate grid. Your child turns that line into the longest side of a right triangle and uses the Pythagorean theorem to find its length. Many classes then write the same steps as one rule, called the distance formula.

What the standard says

The official Common Core wording, quoted exactly. The plain version follows.

8.G.B.8 Apply the Pythagorean Theorem to find the distance between two points in a coordinate system.

What it means, in plain words

The coordinate plane is the grid from algebra class, with a horizontal x-axis and a vertical y-axis that cross at $(0, 0)$ . Each point has an address, called its coordinates, written $(x, y)$ . The point $(4, 6)$ sits 4 units to the right of $(0, 0)$ and 6 units up.

When two points are on the same row or the same column, your child can count the units between them. A slanted line is harder, because the grid squares do not measure it. So your child draws a horizontal line and a vertical line that meet in a square corner. Together with the slanted line they make a right triangle. The slanted line is the hypotenuse (the longest side), and the two shorter sides are the legs.

The coordinates give the legs. The horizontal leg is the difference between the x-coordinates, and the vertical leg is the difference between the y-coordinates. From there the Pythagorean theorem on the coordinate plane works as usual: square the legs, add them and take the square root. The distance formula, $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$ , is the same three steps written in one line.

Your child should be able to:

Find the horizontal and vertical distance between two points from their coordinates
Draw the right triangle that makes a slanted segment its hypotenuse
Use the Pythagorean theorem to find the distance between two points, including points with negative coordinates
Give the answer exactly as a square root or rounded, for example $\sqrt{41} \approx 6.4$

Example

What is the distance between $A(2, 1)$ and $B(8, 9)$ ?

The horizontal leg is $8 - 2 = 6$ units. The vertical leg is $9 - 1 = 8$ units.
$6^2 + 8^2 = 36 + 64 = 100$
$\sqrt{100} = 10$ , so the two points are 10 units apart.

Example

What is the distance between $A(-2, 3)$ and $B(3, -1)$ ? Round to the nearest tenth.

The horizontal leg runs from $x = -2$ to $x = 3$ , so it is $3 - (-2) = 5$ units.
The vertical leg runs from $y = -1$ to $y = 3$ , so it is $3 - (-1) = 4$ units.
$5^2 + 4^2 = 25 + 16 = 41$
$\sqrt{41} \approx 6.4$ , so the two points are about 6.4 units apart.

Where children usually get stuck

For the points $(2, 1)$ and $(8, 9)$ , your child counts 6 squares across and 8 squares up and writes 14 as the distance.
Counting along the grid measures the path around the corner of the triangle. The question asks for the straight line, and the child skipped the step that turns the two legs into a hypotenuse.
What helps. Ask your child to draw both paths on graph paper: the corner path and the straight line. The straight line is visibly shorter than 14, so the answer has to be shorter too.
Your child finds the distance from $x = -2$ to $x = 3$ as $3 - 2 = 1$ unit.
The minus sign on $-2$ gets dropped, so the child subtracts as if both points were on the same side of the y-axis.
What helps. Have your child count the squares on the grid for that one leg. There are 5, which shows that the subtraction has to be $3 - (-2)$ .
For the points $(2, 1)$ and $(8, 9)$ , your child subtracts $9 - 2$ and uses 7 as a leg.
With four numbers on the page, the pairs get mixed up. The child takes an x-coordinate from one point and a y-coordinate from the other.
What helps. Before any subtraction, have your child write "across" above both x-coordinates and "up" above both y-coordinates. Each leg then uses only one kind of number.

Try four practice questions

Real 8th grade questions from our question bank, easiest first. Answer them right here, check each one, and see a worked solution for any question.

1Warm-up
Look at the segment on the grid. Fill in the two blanks with the horizontal and vertical distances.
$\displaystyle ^2 +$
$\displaystyle ^2$
$\displaystyle =$
$\displaystyle d^2$
Show the answer and how to solve it
The blanks are 2 and 3 because those are the horizontal and vertical distances between the segment's endpoints. Writing 3 and 2 also works.
1. 1Use the segment as the hypotenuse. Draw one horizontal leg and one vertical leg to make a right triangle.
2. 2The horizontal leg goes from x = 1 to x = 3, so it is 2 units long. The vertical leg goes from y = 1 to y = 4, so it is 3 units long.
3. 3Put those two leg lengths into the equation. Either order works.
Common mistake. A common mistake is entering 3 and 4 because those are the coordinates of the upper endpoint, not the distances between the two endpoints. Count the spaces between the endpoints along each direction.
The idea underneath. On a coordinate grid, a segment can be modeled as the hypotenuse of a right triangle whose legs are the absolute changes in x and y.
2On level
Look at segment AB on the grid. What is its length? Round your answer to the nearest tenth.
Length of AB:units
Show the answer and how to solve it
Answer. 5.8
The length of segment AB is $5.8$ units because $\sqrt{3^{2} + 5^{2}} = \sqrt{34}$ and $\sqrt{34}$ is about $5.8$ .
1. 1Look at segment AB on the grid. You can form a right triangle using the horizontal and vertical distances between A and B.
2. 2From A to B, the horizontal change is 3 units and the vertical change is 5 units. Both are positive distances; even though A has a negative x-coordinate, distance uses the positive difference of 3 units.
3. 3Let $d$ be the length of AB. Use the Pythagorean theorem with legs 3 and 5.
4. 4Take the square root and round to the nearest tenth. The hundredths digit is 3, so round down.
Common mistake. A common mistake is adding the changes, $3 + 5 = 8$ , instead of squaring each leg first. Another error is using $-1$ as the horizontal change because A has a negative x-coordinate; distance always uses the positive difference of 3 units.
The idea underneath. The segment between two plotted points is the hypotenuse of a right triangle whose legs are the horizontal and vertical changes on the grid.
3On level
What is the exact distance between the two points below?
$\displaystyle A\left(1,3\right)\text{ and }B\left(4,7\right)$
A
$\displaystyle 4$
B
$\displaystyle 5$
C
$\displaystyle 7$
D
$\displaystyle 25$
Show the answer and how to solve it
The answer is 5 because the points are 3 units apart horizontally and 4 units apart vertically, so the straight-line distance is $\sqrt{3^{2} + 4^{2}} = 5$ .
1. 1Look at the ordered pairs. From $A(1,3)$ to $B(4,7)$ , the x-values change by 3 and the y-values change by 4.
2. 2Use those changes in the distance formula.
3. 3Simplify the differences and squares.
4. 4Now take the square root.
Common mistake. A common mistake is stopping at 25 after computing $3^{2} + 4^{2}$ . The distance formula ends with a square root, so $\sqrt{25} = 5$ .
The idea underneath. The distance between two coordinate points is the hypotenuse of a right triangle made from the horizontal and vertical changes.
4Stretch
What is the distance between A and B? Round to the nearest hundredth.
$\displaystyle A\left(-\frac{1}{4}, 0.8\right)\qquad B\left(2.3, -1.5\right)$
A
3.44
B
3.43
C
3.08
D
4.85
Show the answer and how to solve it
The distance is $3.43$ units because $dx = 2.55$ and $dy = 2.3$ , giving $\sqrt{2.55^{2} + 2.3^{2}} = \sqrt{11.7925} \approx 3.4340$ , which rounds to $3.43$ .
1. 1Identify the coordinates of A and B.
2. 2Convert $-\frac{1}{4}$ to $-0.25$ . Then find the horizontal and vertical changes. Subtracting a negative adds.
3. 3Use $dx = 2.55$ and $|dy| = 2.3$ as the legs of a right triangle and apply the distance formula.
4. 4The thousandths digit is $4$ , so $3.4340$ rounds down to $3.43$ . The correct choice is $3.43$ .
Common mistake. A common mistake is treating $2.3 - \left(-\frac{1}{4}\right)$ as $2.3 - 0.25 = 2.05$ . Because subtracting a negative is the same as adding, the horizontal change is $2.55$ , not $2.05$ .
The idea underneath. On the coordinate plane, distance is the hypotenuse of a right triangle whose legs are the horizontal and vertical changes between the points.