Find the Balance Point of Any Triangle by Averaging Its Corners

If a point $P$ lies on the segment joining $A(x_1,y_1)$ and $B(x_2,y_2)$ and divides it in the ratio $m:n$, its coordinates are a weighted average of the ends:

$$P=\left(\frac{m{x}_2+n{x}_1}{m+n},\frac{m{y}_2+n{y}_1}{m+n}\right)$$

When $m=n$, this becomes the mid-point formula. This chapter covers the section formula, finding ratios, mid-points and missing vertices, and the centroid and points of trisection.

**Substitute the ratio $m:n$ and the end points into the section formula, remembering that $m$ multiplies the coordinates of the far end $B$ and $n$ multiplies those of $A$.

Worked example 1.** Find the point dividing $A(2,-3)$ and $B(8,9)$ in the ratio $1:2$.

$$x=\frac{1\times 8+2\times 2}{3}=\frac{12}{3}=4y=\frac{1\times 9+2\times (-3)}{3}=\frac{3}{3}=1$$

The point is $(4,1)$.

Worked example 2. Find the point dividing $A(-1,4)$ and $B(9,-6)$ in the ratio $3:2$.

$$x=\frac{3\times 9+2\times (-1)}{5}=\frac{25}{5}=5y=\frac{3\times (-6)+2\times 4}{5}=\frac{-10}{5}=-2$$

The point is $(5,-2)$.

An everyday example. A toll plaza placed one-third of the way along a straight highway between two towns on a map grid is found with the ratio $1:2$.

The substance. **Swapping $m$ and $n$ gives a different point**: the ratio $1:2$ puts $P$ nearer $A$, while $2:1$ puts it nearer $B$.

**Take the ratio as $k:1$, write one coordinate of the dividing point with the section formula, set it equal to the known value, and solve for $k$; on the x-axis the y-coordinate is $0$, and on the y-axis the x-coordinate is $0$.

Worked example 1 — a given point.** In what ratio does $P(4,1)$ divide $A(2,-3)$ and $B(8,9)$?

$$\frac{8k+2}{k+1}=48k+2=4k+4k=\frac{1}{2}\text{ratio }1:2$$

Worked example 2 — the x-axis. In what ratio does the x-axis divide $A(5,-3)$ and $B(-1,6)$, and where?

$$\frac{6k-3}{k+1}=0k=\frac{1}{2}\text{ratio }1:2$$
$$x=\frac{1\times (-1)+2\times 5}{3}=3\text{point }(3,0)$$

Worked example 3 — the y-axis. For $A(-4,2)$ and $B(6,7)$:

$$\frac{6k-4}{k+1}=0k=\frac{2}{3}\text{ratio }2:3y=\frac{2\times 7+3\times 2}{5}=4\text{point }(0,4)$$

An everyday example. A straight river crossing a road between two villages marked on a grid divides the road in a ratio found exactly like this.

The substance. **Using $k:1$ turns two unknowns into one**, and a positive $k$ confirms the division is internal.

**The mid-point is the average of the end points, $\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)$; reverse it to find a missing end, and use the fact that the diagonals of a parallelogram bisect each other to find a missing vertex.

Worked example 1.** Mid-point of $(-3,5)$ and $(7,-1)$:

$$\left(\frac{-3+7}{2},\frac{5-1}{2}\right)=(2,2)$$

Worked example 2 — missing end. $M(1,4)$ is the mid-point of $A(-2,7)$ and $B$.

$$B=(2\times 1-(-2),2\times 4-7)=(4,1)$$

Worked example 3 — parallelogram. $A(1,2)$, $B(4,3)$, $C(6,6)$ are three vertices of parallelogram $ABCD$. Find $D$.

$$\text{mid-point of }AC=\left(\frac{7}{2},4\right)=\text{mid-point of }BDD=(7-4,8-3)=(3,5)$$

An everyday example. Two friends living on a city grid who agree to meet exactly halfway can find the meeting point by averaging their coordinates.

The substance. The diagonals of a parallelogram bisect each other, so both diagonals share the same mid-point.

**The centroid is the average of the three vertices, $\left(\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3}\right)$, and the points of trisection divide a segment in the ratios $1:2$ and $2:1$.

Worked example 1 — centroid.** $A(2,3)$, $B(-4,1)$, $C(5,-7)$:

$$G=\left(\frac{2-4+5}{3},\frac{3+1-7}{3}\right)=(1,-1)$$

Worked example 2 — missing vertex. The centroid is $G(2,1)$ and two vertices are $A(1,4)$ and $B(5,-2)$.

$$C=(3\times 2-1-5,3\times 1-4+2)=(0,1)$$

Worked example 3 — trisection. Trisect $A(-3,6)$ and $B(9,-3)$.

$$P(1:2)=\left(\frac{9+2(-3)}{3},\frac{-3+2(6)}{3}\right)=(1,3)Q(2:1)=\left(\frac{2(9)-3}{3},\frac{2(-3)+6}{3}\right)=(5,0)$$

An everyday example. A triangular piece of cardboard balances on a pencil tip placed exactly at its centroid.

The substance. **The centroid divides each median in the ratio $2:1$** from the vertex, which is where the formula comes from.