Midpoint and Distance in the Coordinate Plane

By the end of this lesson, you will be able to find the distance between two points using the Distance Formula, locate the midpoint of a segment using the Midpoint Formula, and partition a segment in a given ratio.

The Big Idea

The coordinate plane turns geometric figures into algebra. Instead of measuring with a ruler, you compute. The Distance Formula is the Pythagorean Theorem in disguise; the Midpoint Formula is just averaging coordinates. These two tools are the foundation of all coordinate geometry.

The Distance Formula

For points $(x_1, y_1)$ and $(x_2, y_2)$:

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

Derivation: Draw a right triangle where the segment is the hypotenuse. The horizontal leg has length $|x_2 - x_1|$ and the vertical leg has length $|y_2 - y_1|$. The Pythagorean theorem gives $d^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2$.

The order of subtraction doesn't matter (differences are squared), so $d(A,B) = d(B,A)$.

The Midpoint Formula

The midpoint $M$ of a segment with endpoints $(x_1, y_1)$ and $(x_2, y_2)$:

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

Average the $x$-coordinates; average the $y$-coordinates. The midpoint is always between the two endpoints and equidistant from both.

Example 1 — Distance

Find the distance between $A(-2, 3)$ and $B(4, -1)$.

$$d = \sqrt{(4-(-2))^2 + (-1-3)^2} = \sqrt{36 + 16} = \sqrt{52} = 2\sqrt{13}$$

Example 2 — Midpoint

Find the midpoint of the segment with endpoints $P(3, -5)$ and $Q(-7, 9)$.

$$M = \left(\frac{3 + (-7)}{2},\ \frac{-5 + 9}{2}\right) = \left(\frac{-4}{2},\ \frac{4}{2}\right) = (-2,\ 2)$$

Example 3 — Finding an Endpoint from the Midpoint

The midpoint of $\overline{AB}$ is $M(1, -3)$ and $A = (4, 2)$. Find $B$.

Using the midpoint formula:

$$\frac{4 + x_B}{2} = 1 \implies 4 + x_B = 2 \implies x_B = -2$$

$$\frac{2 + y_B}{2} = -3 \implies 2 + y_B = -6 \implies y_B = -8$$

$B = (-2, -8)$

Verify: Midpoint of $A(4,2)$ and $B(-2,-8)$: $\left(\frac{4-2}{2}, \frac{2-8}{2}\right) = (1, -3)$ ✓

Example 4 — Proving a Figure

Show that the diagonals of the quadrilateral with vertices $A(0,0)$, $B(4,0)$, $C(4,3)$, $D(0,3)$ bisect each other.

Midpoint of diagonal $\overline{AC}$: $\left(\frac{0+4}{2}, \frac{0+3}{2}\right) = (2, 1.5)$

Midpoint of diagonal $\overline{BD}$: $\left(\frac{4+0}{2}, \frac{0+3}{2}\right) = (2, 1.5)$

Same midpoint → diagonals bisect each other. (This figure is a rectangle.) ✓