Chapter 7.2 — Exercise 7.2 — Section Formula

The Construction Behind the Formula

Let P(x,y) divide the segment joining A(x₁,y₁) and B(x₂,y₂) so that AP:PB = m:n. Dropping perpendiculars from A, P, and B down to the x-axis creates two similar triangles — one formed around AP, the other around PB — and comparing their corresponding sides in the same m:n ratio gives both coordinates of P at once.

When m=n — the point sits exactly halfway — the formula collapses into the simpler midpoint formula:

The order of m and n matters just as much as the order of the points themselves: m is always paired with the coordinate of the far point B, and n with the near point A, precisely because P is closer to whichever end has the smaller share of the ratio. Swapping m and n by mistake finds a genuinely different point — the one dividing BA in that ratio instead of AB — so it's worth double-checking which end of the ratio belongs to which point before substituting.

Two Direct Applications

• Point dividing (−1,7) and (4,−3) in ratio 2:3: P = ((2·4+3·−1)/5, (2·−3+3·7)/5) = (5/5, 15/5) = (1, 3).
• Trisection points of (4,−1) and (−2,−3): two points split a segment into three equal parts, one at ratio 1:2 and the other at ratio 2:1 from the same starting point. First: ((2·4+(−2))/3, (2·−1+(−3))/3) = (2, −5/3). Second: ((4+2·−2)/3, (−1+2·−3)/3) = (0, −7/3).

Working Backward to Find the Ratio

Given the dividing point itself, the same formula runs in reverse to recover the ratio it was built from. In what ratio does (−1,6) divide the segment joining (−3,10) and (6,−8)?

Substituting m:n=2:7 into the y-coordinate as a check: (2·−8+7·10)/9 = 54/9 = 6, matching the given point exactly — confirming the ratio without needing to trust the x-equation alone. Checking both coordinates rather than stopping after the first is worth making a habit of specifically for ratio problems, since a single equation in m and n can sometimes be satisfied by an arithmetic slip that the second coordinate would immediately expose.

Diagonals That Bisect Each Other

A parallelogram has one property that turns straight into coordinates: its diagonals always bisect each other, meaning both diagonals share the exact same midpoint. Given (1,2), (4,y), (x,6), (3,5) as the four vertices of a parallelogram taken in order, the midpoint of one diagonal must equal the midpoint of the other:

Solving gives x = 6 and y = 3 — no need to test side lengths or angles at all, since the single midpoint-equality fact settles both unknowns directly. This is a genuinely faster route than measuring all four sides with the distance formula and comparing opposite pairs; recognising which single property actually pins down the unknowns, rather than reaching for the most familiar-looking method, is the real skill this problem is testing.

Reconstructing a Diameter's Missing End

A circle's centre is always the midpoint of any diameter drawn through it. If A is one endpoint of a diameter with centre (2,−3) and the other endpoint is B(1,4), then the midpoint formula, run backward, recovers A directly: (x+1)/2=2 and (y+4)/2=−3, giving A(3, −10). Nothing here needs any circle-specific fact beyond "the centre is the midpoint of every diameter" — the actual arithmetic is the plain midpoint formula solved for an unknown endpoint instead of an unknown midpoint.

A Point Partway Along, Not Halfway

Given A(−2,−2) and B(2,−4), find point P on AB such that AP = 3/7 of the whole length AB. Since AP is 3/7 of AB, the remaining part PB must be the other 4/7 — so AP:PB = 3:4, not 3:7 as it might look at first glance.

Reading the fraction 3/7 as if it were the ratio itself, instead of converting it into the actual AP:PB split first, is the one place this kind of problem most often goes wrong.

Splitting a Segment Into Four Equal Pieces

Finding three points that divide a segment into four equal parts is really just two rounds of the midpoint formula: find the overall midpoint first, then find the midpoint of each half.

P is the midpoint of the first half (from A to Q), and R is the midpoint of the second half (from Q to B) — the same single formula applied three times over rather than any new technique. It's tempting to reach straight for the general section formula with ratios like 1:3 and 3:1 instead, and that would work too, but chaining plain midpoints together is faster here precisely because every one of the four resulting segments is the same length, which is exactly the situation the midpoint formula is built for.

A Point Built Entirely From Letters

The formula works exactly the same way even when every coordinate is itself an expression rather than a number. For (a+b, a−b) and (a−b, a+b), divided internally in ratio 3:2:

Substituting a and b for x₁,y₁,x₂,y₂ exactly as usual, then simplifying the resulting algebraic expression rather than leaving it as a messy fraction of sums, is the only thing that makes this problem different from any of the purely numeric ones above it.

The Centroid: Three Vertices, One Balance Point

A triangle's three medians — each one joining a vertex to the midpoint of the opposite side — always meet at a single point called the centroid, and that point always divides each median in ratio 2:1 measured from the vertex. Averaging all three vertices' coordinates gives the centroid directly, without needing to find even one median explicitly:

Every one of these three triangles has genuinely different-looking vertices, yet the centroid calculation is identical in every row — just three numbers added and divided by three, twice over.

From a Single Point to a Whole Triangle's Area

This exercise found points that sit exactly where a ratio, a midpoint, or a balance-point condition places them. Exercise 7.3 moves from locating single points to measuring an entire enclosed region — the area of a triangle built directly from its three vertices' coordinates, with no need to ever draw the triangle out on paper. For the distance and collinearity ideas this exercise's diagonal- and side-based reasoning leans on, revisit Exercise 7.1.