Lengths and midpoints of line segments

The distance and midpoint formulas, and how to use them to test triangles and quadrilaterals and find circles.

Two formulas

Distance formula

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

Midpoint formula

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

Distance and midpoint

Drag A and B. The length uses the run and the rise; the midpoint averages the coordinates.

Try this

Make the run $3$ and the rise $4$. What is the length?

Example 1●

Find the length and midpoint of the interval from $A(1, 3)$ to $B(7, -1)$.

$AB = \sqrt{(7 - 1)^2 + (-1 - 3)^2} = \sqrt{36 + 16} = \sqrt{52} = 2\sqrt{13}$

$M = \left(\dfrac{1 + 7}{2}, \dfrac{3 + (-1)}{2}\right) = (4, 1)$

Example 2●●

Show that $P(0, 0)$, $Q(4, 1)$, $R(5, 5)$, $S(1, 4)$ form a parallelogram.

Midpoint of $PR = (2.5, 2.5)$

Midpoint of diagonal PR.

Midpoint of $QS = (2.5, 2.5)$

Midpoint of diagonal QS.

The diagonals bisect each other, so $PQRS$ is a parallelogram.

Example 3: a circle●●

$A(-1, 2)$ and $B(5, 10)$ are the ends of a diameter. Find the equation of the circle.

Centre $(2, 6)$

The centre is the midpoint of the diameter.

Radius $= \tfrac12 AB = \tfrac12\sqrt{36 + 64} = 5$

$(x - 2)^2 + (y - 6)^2 = 25$

Flashcards

Distance formula

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

Midpoint formula

Average the $x$-values and the $y$-values.

Parallelogram test

Diagonals have the same midpoint.

Practice

Find the distance from $(1, 2)$ to $(4, 6)$.
The midpoint of $(-2, 5)$ and $(6, -1)$ is:
Find the exact distance from $(1, 3)$ to $(7, -1)$. (Type something like sqrt(20).)
$M(3, 4)$ is the midpoint of $A(-3, -1)$ and $B$. Find the $x$-coordinate of $B$.
To show a quadrilateral is a parallelogram, it is enough to show that:
The triangle with vertices $(0, 0)$, $(4, 0)$ and $(2, 5)$ is:
Find the radius of the circle with diameter from $(0, 0)$ to $(6, 8)$.
The point $(x, 3)$ is $5$ units from $(3, 0)$. Find both possible values of $x$.