Does the order of the points matter?

No. For distance, squaring the difference makes the order irrelevant. For midpoint, addition is commutative, so averaging yields the same result regardless of order.

What if the points share an $x$- or $y$-coordinate?

The distance formula simplifies to the absolute value of the difference of the other coordinate. The midpoint formula remains exactly the same.

Can this be extended to 3D space?

Yes. Add the $z$-coordinate. The distance becomes $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}$ and the midpoint averages all three sets of coordinates.

How to find the distance and midpoint between two points

The distance between two points is found using the Pythagorean theorem applied to their coordinates, while the midpoint is the average of their respective $x$ - and $y$ -coordinates. This method applies whenever points are defined in a standard 2D Cartesian coordinate system.

The setup

Define the two given points as $P_1 = (x_1, y_1)$ and $P_2 = (x_2, y_2)$ . The distance $d$ between them is calculated using the formula $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$ . The midpoint $M$ of the segment connecting them is calculated using the formula $M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)$ .

The steps

Identify and label the coordinates of $P_1$ and $P_2$ . 2. Substitute the $x$ - and $y$ -values into the distance formula, square the differences, sum them, and take the square root. 3. Substitute the values into the midpoint formula, add the respective coordinates, and divide by 2.

Checking the result

Verify that the calculated distance is strictly non-negative. To check the midpoint, confirm it lies exactly on the line segment by checking if the distance from $P_1$ to $M$ equals the distance from $M$ to $P_2$ , and that their sum equals the total distance $d$ .

Common errors

A frequent error is subtracting coordinates in the midpoint formula instead of adding them, confusing it with the slope or distance formula. Another error is adding coordinates in the distance formula before squaring, rather than subtracting.

Worked example

Find the distance and midpoint between $A = (-3, 4)$ and $B = (5, -2)$ .

Let $(x_1, y_1) = (-3, 4)$ and $(x_2, y_2) = (5, -2)$ . Calculate distance: $d = \sqrt{(5 - (-3))^2 + (-2 - 4)^2}$ $d = \sqrt{(8)^2 + (-6)^2}$ $d = \sqrt{64 + 36}$ $d = \sqrt{100}$ $d = 10$ Calculate midpoint: $M = \left( \frac{-3 + 5}{2}, \frac{4 + (-2)}{2} \right)$ $M = \left( \frac{2}{2}, \frac{2}{2} \right)$ $M = (1, 1)$ . The distance is $10$ and the midpoint is $(1, 1)$ .

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References: OpenStax College Algebra, Chapter 2: Equations and Inequalities · Stewart Calculus, Appendix B: Coordinate Geometry