High School Geometry - Distance and Midpoint Formulas

Distance Formula

If the coordinates of any two points are known, then the distance between the points can easily be found using the distance formula.
Consider two points $P (x_{1}, y_{1})$ and $Q (x_{2}, y_{2})$ as shown on the graph below.

$|P Q| = \sqrt{{(∆ a b s c i s s a)}^{2} + {(∆ o r d i n a t e)}^{2}}$

$= \sqrt{{(x_{1} - x_{2})}^{2} + {(y_{1} - y_{2})}^{2}}$ , or

$= \sqrt{{(x_{2} - x_{1})}^{2} + {(y_{2} - y_{1})}^{2}}$

If the coordinates of the endpoints of a line segment are known, then the coordinates of its midpoint can easily be found using the midpoint formula.
Consider a line segment AB as shown in the figure.

Let M be the mid-point of AB, then the coordinates of M are given by the formula:

$M (x, y) = (\frac{x_{1} + x_{2}}{2}, \frac{y_{1} + y_{2}}{2})$

Question 1: Find the distance between the points P and Q shown in the graph.

Solution: It is given that P(1, 2) & Q(4, 6).

$P (1, 2) \to x_{1} = 1 & y_{1} = 2$

$Q (4, 6) \to x_{2} = 4 & y_{2} = 6$

$|P Q| = \sqrt{{(x_{2} - x_{1})}^{2} + {(y_{2} - y_{1})}^{2}}$

$= \sqrt{{(4 - 1)}^{2} + {(6 - 2)}^{2}}$

$= \sqrt{9 + 16} = \sqrt{25} = 5$

Question 2: Find the midpoint of the line segment AB shown in the graph.

Solution: It is given that A(-4, 3) & B(5, 1).

$A (- 4, 3) \to x_{1} = - 4 & y_{1} = 3$

$B (5, 1) \to x_{2} = 5 & y_{2} = 1$

Let $M (x, y)$ be the coordinate of the midpoint of AB.

$x = \frac{x_{1} + x_{2}}{2} = \frac{- 4 + 5}{2} = \frac{1}{2}$

$y = \frac{y_{1} + y_{2}}{2} = \frac{3 + 1}{2} = 2$

Hence, the midpoint of segment AB is $M (\frac{1}{2}, 2)$ .

The distance between the points $P (x_{1}, y_{1})$ and $Q (x_{2}, y_{2})$ is given by:

$|P Q| = \sqrt{{(x_{2} - x_{1})}^{2} + {(y_{2} - y_{1})}^{2}}$

The coordinates of the midpoint of a line segment AB where $A (x_{1}, y_{1})$ and $B (x_{2}, y_{2})$ are given by:

$M (x, y) = (\frac{x_{1} + x_{2}}{2}, \frac{y_{1} + y_{2}}{2})$