The section formula and midpoint

If $P$ and $Q$ are two points and we want to find the point that divides $PQ$ in a given ratio, the section formula does the job in one shot.

Statement

Let $P = (x_1, y_1)$ and $Q = (x_2, y_2)$ , and let $R$ be a point on segment $PQ$ such that $PR : RQ = m : n$ . Then $\boxed{R = \left(\frac{m x_2 + n x_1}{m + n}, \frac{m y_2 + n y_1}{m + n}\right).}$

Special case: if $R$ is the midpoint ( $m : n = 1 : 1$ ), $R = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right).$

Derivation

Draw the segment from $P$ to $Q$ . By similar triangles (drop perpendiculars to the $x$ - and $y$ -axes), the $x$ -coordinate of $R$ satisfies $\frac{R_x - x_1}{x_2 - R_x} = \frac{PR}{RQ} = \frac{m}{n}.$ Solving: $n (R_x - x_1) = m (x_2 - R_x) \Rightarrow R_x (m + n) = m x_2 + n x_1 \Rightarrow R_x = \dfrac{m x_2 + n x_1}{m + n}$ .

A parallel argument with $y$ -coordinates gives the second component. $\blacksquare$

When to use

Trisection points of a segment: $PR : RQ = 1 : 2$ for the first trisector, $2 : 1$ for the second.
Midpoint of any segment.
Centroid of a triangle: average of the three vertices: $\left(\dfrac{x_1 + x_2 + x_3}{3}, \dfrac{y_1 + y_2 + y_3}{3}\right)$ .
A point dividing a segment in a known ratio (often given in problems).

A useful inverse: if you know $P, Q,$ and $R$ (on segment $PQ$ ), you can compute the ratio $PR : RQ$ from coordinates. Just rearrange.

Worked examples

Example 1. Find the midpoint of $A = (2, 7)$ and $B = (-4, 5)$ .

Midpoint $= \left(\dfrac{2 - 4}{2}, \dfrac{7 + 5}{2}\right) = (-1, 6)$ .

Example 2. Find the point dividing the segment from $A = (1, 2)$ to $B = (6, 7)$ in the ratio $2 : 3$ .

$R = \left(\dfrac{2 \cdot 6 + 3 \cdot 1}{2 + 3}, \dfrac{2 \cdot 7 + 3 \cdot 2}{2 + 3}\right) = \left(\dfrac{15}{5}, \dfrac{20}{5}\right) = (3, 4)$ .

Example 3. Find the trisection points of the segment from $A = (4, -1)$ to $B = (-2, -3)$ .

Trisection points divide the segment in ratios $1 : 2$ and $2 : 1$ .

$P = \left(\dfrac{1 \cdot (-2) + 2 \cdot 4}{3}, \dfrac{1 \cdot (-3) + 2 \cdot (-1)}{3}\right) = \left(\dfrac{6}{3}, \dfrac{-5}{3}\right) = (2, -5/3)$ .

$Q = \left(\dfrac{2 \cdot (-2) + 1 \cdot 4}{3}, \dfrac{2 \cdot (-3) + 1 \cdot (-1)}{3}\right) = \left(\dfrac{0}{3}, \dfrac{-7}{3}\right) = (0, -7/3)$ .

Example 4. Find the ratio in which the $x$ -axis divides the segment from $A = (-2, 3)$ to $B = (5, -4)$ .

A point on the $x$ -axis has $y = 0$ . Suppose it divides $AB$ in ratio $m : 1$ (call it $k : 1$ ). The $y$ -coordinate of the divider is $\frac{k \cdot (-4) + 1 \cdot 3}{k + 1} = 0 \Rightarrow -4 k + 3 = 0 \Rightarrow k = 3/4.$ So the ratio is $3 : 4$ .

Example 5. Find the centroid of the triangle with vertices $(2, 3), (-1, 4), (5, -1)$ .

Centroid $= \left(\dfrac{2 - 1 + 5}{3}, \dfrac{3 + 4 - 1}{3}\right) = (2, 2)$ .

Try it yourself

Midpoint of $(3, -7)$ and $(-1, 5)$ .
Point dividing $A = (-3, 5)$ to $B = (4, -2)$ in ratio $2 : 3$ .
Find both trisection points of segment from $(1, -2)$ to $(4, 1)$ .
Find the ratio in which the $y$ -axis divides the segment from $(-3, 5)$ to $(4, -2)$ .
Find the centroid of the triangle with vertices $(3, -5), (-7, 4), (10, -2)$ .
The midpoint of $AB$ is $(2, -3)$ ; if $A = (5, 1)$ , find $B$ .
Find $k$ if $(k, 4)$ is the midpoint of $(2, 5)$ and $(8, k - 1)$ .
Find the coordinates of the point that divides the segment from $(3, 0)$ to $(-5, 8)$ externally in ratio $3 : 5$ . (For external division use $m x_2 - n x_1 / (m - n)$ .)
If $(a, b)$ is the midpoint of $(2a, 3b)$ and $(c, d)$ , find $c, d$ in terms of $a, b$ .
In what ratio is the segment joining $(2, -3)$ and $(5, 6)$ divided by the $x$ -axis?

Pitfalls / Insight

The ratio $m : n$ corresponds to $m x_2 + n x_1$ , not $m x_1 + n x_2$ . The number in the numerator near $x_2$ is the near- $Q$ part of the ratio.
For the midpoint formula, just take the average. No fractions to mix up.
Negative coordinates can confuse signs. Compute carefully.

Insight. The section formula is a weighted average of two points. Bigger weight on $Q$ pulls the result closer to $Q$ . Internalise it and you'll spot ratios in many geometry problems.