Distance Formula

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The distance formula finds the length of the segment between two points on a coordinate plane. For (x_1, y_1) and (x_2, y_2), the distance is d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}. Subtract, square, add, then take the square root.

Before this lesson: You should know the Midpoint Formula, the Coordinate Plane and Simplifying Square Roots.

Where the Distance Formula Comes From

Pick two points that are not on the same row or column. Draw a horizontal leg and a vertical leg between them. Now the segment is the long side of a right triangle.

The horizontal leg is x_2 - x_1 units long. The vertical leg is y_2 - y_1 units long. The Pythagorean theorem says a^2 + b^2 = c^2 for a right triangle. So the segment’s length squared is the sum of the legs squared.

A coordinate grid from 0 to 7 across and 0 to 7 up. Points A(1, 2) and B(4, 6) are joined by a blue segment labeled d = 5. A green dashed horizontal leg from A to C(4, 2) is labeled 3, and a green dashed vertical leg from C to B is labeled 4. A small square marks the right angle at C. Beside the graph: 3² + 4² = 9 + 16 = 25, and the square root of 25 is 5.
The legs are 3 and 4, so the distance is the square root of 9 + 16 = 25, which is 5.

Take the square root of both sides, and you get the distance formula. A distance is never negative, so we only use the positive square root.

The Distance Formula

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

Squaring makes each difference positive. So it doesn’t matter which point you call the first one.

Steps to use the distance formula

  1. Label the points (x_1, y_1) and (x_2, y_2).
  2. Subtract the x-coordinates. Subtract the y-coordinates.
  3. Square each difference.
  4. Add the squares.
  5. Take the square root. Simplify the radical if it is not a whole number.

Worked Examples

Example 1: A whole-number distance

Find the distance between A(1, 2) and B(4, 6).

Step 1: Subtract the x-coordinates.

4 - 1 = 3

Step 2: Subtract the y-coordinates.

6 - 2 = 4

Step 3: Put the differences in the formula.

d = \sqrt{3^2 + 4^2}

Step 4: Square each difference.

d = \sqrt{9 + 16}

Step 5: Add.

d = \sqrt{25}

Step 6: Take the square root.

d = 5

The distance is 5 units. The graph above shows this triangle.

Example 2: Negative coordinates

Find the distance between P(-2, 3) and Q(4, -5).

Step 1: Subtract the x-coordinates. Subtracting a negative means adding (see adding and subtracting integers).

4 - (-2) = 6

Step 2: Subtract the y-coordinates.

-5 - 3 = -8

Step 3: Put the differences in the formula.

d = \sqrt{6^2 + (-8)^2}

Step 4: Square each difference. A negative number squared is positive.

d = \sqrt{36 + 64}

Step 5: Add.

d = \sqrt{100}

Step 6: Take the square root.

d = 10

The distance is 10 units.

Example 3: Simplify the radical

Find the distance between C(-1, -4) and D(3, 2). Give the exact answer.

Step 1: Subtract the x-coordinates.

3 - (-1) = 4

Step 2: Subtract the y-coordinates.

2 - (-4) = 6

Step 3: Put the differences in the formula.

d = \sqrt{4^2 + 6^2}

Step 4: Square each difference.

d = \sqrt{16 + 36}

Step 5: Add.

d = \sqrt{52}

Step 6: 52 is not a perfect square. Find its largest perfect square factor (see simplifying square roots).

d = \sqrt{4 \cdot 13}

Step 7: Take the square root of 4.

d = 2\sqrt{13}

The exact distance is 2\sqrt{13} units, or about 7.2 units.

Example 4: Classify a triangle by its sides

A triangle has corners A(1, 1), B(4, 5) and C(8, 2). Is it isosceles? (An isosceles triangle has at least two sides of equal length.)

Step 1: Find AB. The differences are 4 - 1 = 3 and 5 - 1 = 4.

AB = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5

Step 2: Find BC. The differences are 8 - 4 = 4 and 2 - 5 = -3.

BC = \sqrt{4^2 + (-3)^2} = \sqrt{16 + 9} = \sqrt{25} = 5

Step 3: Find AC. The differences are 8 - 1 = 7 and 2 - 1 = 1.

AC = \sqrt{7^2 + 1^2} = \sqrt{49 + 1} = \sqrt{50}

Step 4: Simplify \sqrt{50}.

AC = \sqrt{25 \cdot 2} = 5\sqrt{2}

Step 5: Compare the sides.

AB = BC = 5

Two sides are equal, so the triangle is isosceles. \overline{AB} \cong \overline{BC}.

A coordinate grid from 0 to 9 across and 0 to 7 up. Triangle ABC has corners A(1, 1), B(4, 5) and C(8, 2). Sides AB and BC are blue and each has one tick mark and the label 5. Side AC is orange and dashed, labeled 5√2.
AB and BC are both 5 units long, so triangle ABC is isosceles.

Example 5: Word problem

A drone starts at (2, -1) on a field map. It flies in a straight line to (14, 4). Each unit on the map is 1 kilometer. How far does it fly?

Step 1: Subtract the x-coordinates.

14 - 2 = 12

Step 2: Subtract the y-coordinates.

4 - (-1) = 5

Step 3: Put the differences in the formula.

d = \sqrt{12^2 + 5^2}

Step 4: Square each difference.

d = \sqrt{144 + 25}

Step 5: Add.

d = \sqrt{169}

Step 6: Take the square root.

d = 13

The drone flies 13 kilometers.

Common Mistakes

MistakeExample of the errorFix
Adding the coordinatesUsing 1 + 4 for A(1, 2) and B(4, 6)Subtract the coordinates: 4 - 1 = 3. Adding is for the midpoint.
Mixing x and yUsing x_2 - y_1Subtract x from x and y from y.
Sign errors with negativesWriting 4 - (-2) = 2Subtracting a negative means adding: 4 + 2 = 6.
Taking the root of each partWriting \sqrt{9 + 16} = 3 + 4 = 7Add first: \sqrt{25} = 5. The root of a sum is not the sum of the roots.
Squaring a negative wrongWriting (-8)^2 = -64(-8)^2 = 64. A square is never negative.
Not simplifying the radicalLeaving \sqrt{52}Factor out the perfect square: \sqrt{52} = 2\sqrt{13}.

Horizontal, Vertical and Number Line Distances

When two points are on the same horizontal line, the y-coordinates match. Then the distance is just the difference of the x-coordinates. For (-3, 2) and (5, 2), the distance is |5 - (-3)| = 8.

When two points are on the same vertical line, use the y-coordinates. For (1, -4) and (1, 6), the distance is |6 - (-4)| = 10.

On a number line, the distance is |a - b|, as in the segment addition postulate. The distance formula gives the same answers, because one of the squares is 0.

FormulaWhat it findsOperation
Midpoint formulaThe point halfway betweenAdd, then divide by 2
Slope formulaHow steep the segment isSubtract, then divide
Distance formulaHow long the segment isSubtract, square, add, square root

Practice Problems

  1. Find the distance between (0, 0) and (6, 8).
  2. Find the distance between (2, 3) and (5, 7).
  3. Find the distance between (-4, 1) and (2, 9).
  4. Find the distance between (3, -2) and (3, 7).
  5. Find the distance between (-5, -1) and (7, 4).
  6. Find the exact distance between (1, 2) and (4, 5).
  7. Find the exact distance between (-2, -3) and (2, 5).
  8. Find the exact distance between (-1, 4) and (5, 1).
  9. A triangle has corners P(0, 0), Q(4, 0) and R(2, 5). Is it isosceles?
  10. Two hikers are at (-3, -2) and (5, 13) on a trail map. Each unit is 1 kilometer. How far apart are they in a straight line?
Show answers
  1. Subtract the x-coordinates: 6 - 0 = 6. Subtract the y-coordinates: 8 - 0 = 8. Square each difference: 6^2 + 8^2 = 36 + 64. Add: 36 + 64 = 100. Take the square root: \sqrt{100} = 10. The distance is 10.
  2. Subtract the x-coordinates: 5 - 2 = 3. Subtract the y-coordinates: 7 - 3 = 4. Square each difference: 3^2 + 4^2 = 9 + 16. Add: 9 + 16 = 25. Take the square root: \sqrt{25} = 5. The distance is 5.
  3. Subtract the x-coordinates: 2 - (-4) = 6. Subtract the y-coordinates: 9 - 1 = 8. Square each difference: 6^2 + 8^2 = 36 + 64. Add: 36 + 64 = 100. Take the square root: \sqrt{100} = 10. The distance is 10.
  4. The x-coordinates match, so the points are on a vertical line. Subtract the y-coordinates: 7 - (-2) = 9. The distance is 9. (The formula gives \sqrt{0^2 + 9^2} = \sqrt{81} = 9.)
  5. Subtract the x-coordinates: 7 - (-5) = 12. Subtract the y-coordinates: 4 - (-1) = 5. Square each difference: 12^2 + 5^2 = 144 + 25. Add: 144 + 25 = 169. Take the square root: \sqrt{169} = 13. The distance is 13.
  6. Subtract the x-coordinates: 4 - 1 = 3. Subtract the y-coordinates: 5 - 2 = 3. Square each difference: 3^2 + 3^2 = 9 + 9. Add: 9 + 9 = 18. Take the square root: \sqrt{18}. Simplify: \sqrt{9 \cdot 2} = 3\sqrt{2}. The distance is 3\sqrt{2}.
  7. Subtract the x-coordinates: 2 - (-2) = 4. Subtract the y-coordinates: 5 - (-3) = 8. Square each difference: 4^2 + 8^2 = 16 + 64. Add: 16 + 64 = 80. Take the square root: \sqrt{80}. Simplify: \sqrt{16 \cdot 5} = 4\sqrt{5}. The distance is 4\sqrt{5}.
  8. Subtract the x-coordinates: 5 - (-1) = 6. Subtract the y-coordinates: 1 - 4 = -3. Square each difference: 6^2 + (-3)^2 = 36 + 9. Add: 36 + 9 = 45. Take the square root: \sqrt{45}. Simplify: \sqrt{9 \cdot 5} = 3\sqrt{5}. The distance is 3\sqrt{5}.
  9. Find PQ. The points are on the x-axis: PQ = 4 - 0 = 4. Find PR: PR = \sqrt{(2 - 0)^2 + (5 - 0)^2} = \sqrt{4 + 25} = \sqrt{29}. Find QR: QR = \sqrt{(2 - 4)^2 + (5 - 0)^2} = \sqrt{4 + 25} = \sqrt{29}. Compare: PR = QR. Yes. Two sides are \sqrt{29}, so the triangle is isosceles.
  10. Subtract the x-coordinates: 5 - (-3) = 8. Subtract the y-coordinates: 13 - (-2) = 15. Square each difference: 8^2 + 15^2 = 64 + 225. Add: 64 + 225 = 289. Take the square root: \sqrt{289} = 17. The hikers are 17 kilometers apart.

FAQ

What is the distance formula?

d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}. It gives the length of the segment between two points.

Is the distance formula the same as the Pythagorean theorem?

Yes, in a way. The two differences are the legs of a right triangle, and the distance is the long side. The formula is the Pythagorean theorem written with coordinates.

Does the order of the points matter?

No. Each difference gets squared, and a number and its opposite have the same square.

Can the distance be negative?

No. A distance is a length, so it is always 0 or more. That is why we take only the positive square root.

Summary

  • Distance formula: d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}.
  • Subtract, square, add, then take the square root.
  • The formula comes from a right triangle whose legs are the x and y differences.
  • Simplify radicals for exact answers, like \sqrt{52} = 2\sqrt{13}.
  • For points on a horizontal or vertical line, just subtract the coordinates that change.

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