📍Midpoint & Distance

Mark two points on a map and two questions appear at once: where is the middle between them, and how far apart are they? On the grid of coordinates, both questions have clean formulas — one uses averaging, the other uses the Pythagorean theorem.

Midpoint: average the coordinates

The midpoint of two points takes the average of the x-coordinates and the average of the y-coordinates:

$\left( \frac{x_1 + x_2}{2},\ \frac{y_1 + y_2}{2} \right)$

For $(2, 3)$ and $(6, 7)$: the x-coordinate is $\frac{2+6}{2} = 4$, the y-coordinate is $\frac{3+7}{2} = 5$, so the midpoint is $(4, 5)$. Why average? Because "dead center" means equally far from both ends — and the average stands exactly in the middle of two numbers.

Measure the flat cases first

Distance comes free in two special situations:

• Same row (equal y-coordinates): the distance is the difference of the x-coordinates, e.g. $(1, 2)$ to $(5, 2)$ is 4 apart;
• Same column (equal x-coordinates): the distance is the difference of the y-coordinates.

Slanted pairs are not so generous — but they do hand over these two "legs": from $(2,3)$ to $(6,7)$ the horizontal gap is $6-2=4$ and the vertical gap is $7-3=4$.

Slanted distance: Pythagoras steps in

Treat the horizontal and vertical gaps as the two legs of a right triangle; the slanted road is the hypotenuse. From $(0,0)$ to $(3,4)$: go 3 across and 4 up, so

$d = \sqrt{3^2 + 4^2} = \sqrt{25} = 5$

The classic 3-4-5 triangle. Any slanted distance can be measured by assembling an invisible right triangle like this.

The distance formula: Pythagoras in disguise

Write the routine as a general rule and you get the distance formula:

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

It is not a new theorem — the first difference is the horizontal leg, the second is the vertical leg, and it is still Pythagoras wearing a coat. Applied to $(2,3)$ and $(6,7)$: $d = \sqrt{4^2 + 4^2} = \sqrt{32} \approx 5.7$.

x3y-2

(3, -2) → Quadrant IV

Walk x steps horizontally, then y steps vertically — the ordered pair (x, y) pins down exactly one point.

Drag the points on the plane: see how easy distance is on the same row or column, then slant them and feel the moment Pythagoras takes over.

Check yourself

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Quick quiz

↺ Try again

1. 1. What is the midpoint of (1, 2) and (5, 6)?

   (3, 4)(6, 8)(2, 3)

2. 2. What is the distance from (0, 0) to (6, 8)?

   10147

3. 3. What is the distance from (0, 0) to (5, 12)?

   13178.5