Distance and Section Formula

Given two points in the coordinate plane, we often need to find the distance between them or to locate a point that divides the segment between them in a prescribed ratio. Both tasks reduce to straightforward formulas derived from the Pythagorean theorem.

Distance Formula

⚡Theorem — Distance Formula (Theorem 9.1)

The distance between points $A(x_A,\, y_A)$ and $B(x_B,\, y_B)$ is

$\boxed{|AB| = \sqrt{(x_B - x_A)^2 + (y_B - y_A)^2}}$

Proof. Drop perpendiculars from $A$ and $B$ to the coordinate axes to form a right triangle $ACB$ , where $C = (x_B, y_A)$ is the corner point. The legs of this right triangle have lengths $|x_B - x_A|$ (horizontal) and $|y_B - y_A|$ (vertical). By the Pythagorean theorem:

$|AB|^2 = |AC|^2 + |CB|^2 = (x_B - x_A)^2 + (y_B - y_A)^2$

Taking the positive square root gives the formula. $\blacktriangleleft$

Special cases:

If $A$ and $B$ lie on a horizontal line ( $y_A = y_B$ ), then $|AB| = |x_B - x_A|$ .
If $A$ and $B$ lie on a vertical line ( $x_A = x_B$ ), then $|AB| = |y_B - y_A|$ .
The distance from the origin $O(0,0)$ to a point $P(x,y)$ is $|OP| = \sqrt{x^2 + y^2}$ .

Midpoint Formula

A direct corollary of the section formula (below) is the midpoint formula: if $M$ is the midpoint of segment $AB$ , then

$\boxed{M = \left(\frac{x_A + x_B}{2},\; \frac{y_A + y_B}{2}\right)}$

Proof. A midpoint divides the segment in ratio $1:1$ . Setting $m = n = 1$ in the section formula gives this result immediately. $\blacktriangleleft$

Section (Division) Formula

⚡Theorem — Section Formula (Theorem 9.2)

Let $C$ be a point on segment $AB$ (between $A$ and $B$ ) that divides it internally in the ratio $AC : CB = m : n$ , where $m, n > 0$ . Then the coordinates of $C$ are:

$\boxed{x_C = \frac{n\, x_A + m\, x_B}{m + n}, \qquad y_C = \frac{n\, y_A + m\, y_B}{m + n}}$

Proof. Since $C$ lies on segment $AB$ with $AC : CB = m : n$ , the parameter $t = \dfrac{m}{m+n}$ gives the fraction of the way from $A$ to $B$ . The coordinates of $C$ are therefore:

$x_C = x_A + t(x_B - x_A) = x_A + \frac{m}{m+n}(x_B - x_A) = \frac{(m+n)x_A + m(x_B - x_A)}{m+n} = \frac{n\,x_A + m\,x_B}{m+n}$

An identical argument applies to $y_C$ . $\blacktriangleleft$

Memory aid: The coordinate corresponding to endpoint $A$ carries weight $n$ (the ratio part near $B$ ), and vice versa — the numerator “crosses over.”

Worked Examples

✎Example — Example 1 — Distance Between Two Points

Find the distance between $P(-3, 4)$ and $Q(5, -2)$ .

Solution. Apply the distance formula with $x_A = -3$ , $y_A = 4$ , $x_B = 5$ , $y_B = -2$ :

$|PQ| = \sqrt{(5 - (-3))^2 + (-2 - 4)^2} = \sqrt{8^2 + (-6)^2} = \sqrt{64 + 36} = \sqrt{100} = 10$

Answer: $|PQ| = 10$ . $\blacktriangleleft$

✎Example — Example 2 — Section Formula

Point $C$ divides segment $AB$ , where $A(1, -1)$ and $B(7, 5)$ , in the ratio $AC : CB = 2 : 1$ . Find the coordinates of $C$ .

Solution. Here $m = 2$ , $n = 1$ . Apply the section formula:

$x_C = \frac{n \cdot x_A + m \cdot x_B}{m + n} = \frac{1 \cdot 1 + 2 \cdot 7}{2 + 1} = \frac{1 + 14}{3} = \frac{15}{3} = 5$

$y_C = \frac{n \cdot y_A + m \cdot y_B}{m + n} = \frac{1 \cdot (-1) + 2 \cdot 5}{3} = \frac{-1 + 10}{3} = \frac{9}{3} = 3$

Verification: $|AC| = \sqrt{(5-1)^2 + (3-(-1))^2} = \sqrt{16+16} = 4\sqrt{2}$ and $|CB| = \sqrt{(7-5)^2+(5-3)^2} = 2\sqrt{2}$ . Indeed $|AC|:|CB| = 2:1$ . $\checkmark$

Answer: $C(5,\, 3)$ . $\blacktriangleleft$

✎Example — Example 3 — Midpoint Application

The midpoint of segment $PQ$ is $M(2, -3)$ . If $P = (-4, 1)$ , find $Q$ .

Solution. Let $Q = (x, y)$ . By the midpoint formula:

$\frac{-4 + x}{2} = 2 \implies x = 8, \qquad \frac{1 + y}{2} = -3 \implies y = -7$

Answer: $Q(8,\, -7)$ . $\blacktriangleleft$

Exercises

✏Exercise

Find the distance between each pair of points.

$A(0, 0)$ and $B(5, 12)$
$A(-2, 3)$ and $B(4, -5)$
$A(1, 1)$ and $B(4, 5)$

Hint for (3): Form the right triangle and apply the Pythagorean theorem directly to check your answer.

✏Exercise

Point $D$ divides segment $EF$ , where $E(-5, 2)$ and $F(7, -4)$ , in the ratio $ED : DF = 1 : 3$ . Find the coordinates of $D$ .

Then verify your answer by computing $|ED|$ and $|DF|$ and checking their ratio.

✏Exercise

Three vertices of a rhombus are $A(0, 0)$ , $B(4, 0)$ , and $C(5, 3)$ . Find the fourth vertex $D$ , given that $ABCD$ is a rhombus (diagonals of a parallelogram bisect each other).

Hint: Use the midpoint formula — the diagonals of a parallelogram share the same midpoint.