The Distance Between Two Points

Straight across the grid

How far is it from (1, 1) to (4, 5) in a straight line? Counting steps along the grid lines, 3 across and then 4 up, gives 3 + 4 = 7. But that is the length of a path with a corner in it. The straight line between the points is shorter.

A right triangle under the line

Go from (1, 1) across to (4, 1), then up to (4, 5). The first move is horizontal and the second is vertical, so they meet at a right angle at (4, 1). Together with the straight line from (1, 1) to (4, 5), they make a right triangle.

The two shorter sides are the run, 4 − 1 = 3, and the rise, 5 − 1 = 4. The straight line between the two points is the side opposite the right angle: the hypotenuse.

The run from (1, 1) to (4, 1) is 3 and the rise from (4, 1) to (4, 5) is 4. They meet at a right angle, and the line from (1, 1) to (4, 5) is the hypotenuse.

Pythagoras on the grid

In a right triangle, the square of the hypotenuse is the sum of the squares of the two shorter sides. So the distance d satisfies $d^2=3^2+4^2=9+16=25$d² = 3² + 4² = 9 + 16 = 25, and $d=\sqrt{25}=5$d = √25 = 5.

The straight line is 5 long, and the path along the grid lines is 7. The hypotenuse is always shorter than the two other sides added together.

A and B sit at opposite corners of the dashed box, whose sides are the run and the rise. The distance AB is the diagonal of the box, the hypotenuse of a right triangle with those two sides, and the working at the top left finds it by Pythagoras’ theorem. When the box is 3 by 4, $AB=\sqrt{9+16}=5$AB = √(9 + 16) = 5.

The distance formula

For any two points $\left(x_1,y_1\right)$(x₁, y₁) and $\left(x_2,y_2\right)$(x₂, y₂), the run is $x_2-x_1$x₂ − x₁ and the rise is $y_2-y_1$y₂ − y₁. Square them, add, and take the square root: $d=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}$d = √((x₂ − x₁)² + (y₂ − y₁)²).

From (−2, 3) to (4, −5), the run is 4 − (−2) = 6 and the rise is −5 − 3 = −8. Then $d=\sqrt{6^2+{\left(-8\right)}^2}=\sqrt{36+64}=\sqrt{100}=10$d = √(6² + (−8)²) = √(36 + 64) = √100 = 10.

The rise came out negative, because the second point is lower. That does not matter: the whole difference is squared, and ${\left(-8\right)}^2=64$(−8)² = 64, the same as $8^2$8². So it makes no difference which point is taken first.

From (−2, 3) to (4, −5): 6 across and 8 down. The triangle has shorter sides 6 and 8, so the distance is $\sqrt{36+64}=10$√(36 + 64) = 10.

When the root is not a whole number

From (1, 2) to (4, 4), the run is 3 and the rise is 2, so $d=\sqrt{9+4}=\sqrt{13}$d = √(9 + 4) = √13. No whole number squares to 13, so the exact distance is $\sqrt{13}$√13, which is about 3.61. Leave it as $\sqrt{13}$√13 unless a decimal is asked for.

A root can sometimes be simplified. From (0, 0) to (2, 4), $d=\sqrt{4+16}=\sqrt{20}$d = √(4 + 16) = √20, and $\sqrt{20}=\sqrt{4\times 5}=2\sqrt{5}$√20 = √(4 × 5) = 2√5.

To compare two distances, compare their squares. Is Q(4, 6) or R(7, 2) nearer to P(1, 1)? PQ² $=3^2+5^2=9+25=34$= 3² + 5² = 9 + 25 = 34, and PR² $=6^2+1^2=36+1=37$= 6² + 1² = 36 + 1 = 37. The smaller square belongs to the shorter distance, so Q is nearer, and no square root had to be worked out.

Points level with each other

When two points have the same y-coordinate, there is no rise, and the distance is the difference in x. From (2, 3) to (9, 3) it is 9 − 2 = 7. The formula agrees: $\sqrt{7^2+0^2}=\sqrt{49}=7$√(7² + 0²) = √49 = 7. Points with the same x-coordinate work the same way, with the difference in y.

The usual mistakes

Adding the run and the rise. For (1, 1) and (4, 5), 3 + 4 = 7 is the path along the grid lines. The straight-line distance is $\sqrt{3^2+4^2}=5$√(3² + 4²) = 5.

Stopping at the square. $3^2+4^2=25$3² + 4² = 25 is $d^2$d², not d. The last step is the square root: $d=\sqrt{25}=5$d = √25 = 5.

Squaring a negative difference without its bracket. $-8^2$−8² read as −64 gives $\sqrt{36-64}$√(36 − 64), which has no answer. The difference is −8, and ${\left(-8\right)}^2=64$(−8)² = 64.