Give every point an address, and geometry starts keeping its books with algebra.
For two thousand years geometry and algebra lived in separate rooms — one drew pictures, the other pushed symbols. Then René Descartes laid a shape on grid paper and gave every point an address, a pair of numbers (x, y). Suddenly a figure was just a set of addresses, and a geometric question — how far apart? where is the middle? — became arithmetic you could simply compute. A distance turns into a square root, a midpoint into an average, a direction into an arrow. This lesson is where geometry first opens its account book and starts keeping its records in algebra. (If the plane itself feels new, revisit the coordinate plane first.)
Here is the whole move in one sentence. Lay down two number lines that cross at right angles — the x-axis and the y-axis — and every point in the plane now has a unique address (x, y): how far across, then how far up. The point is the pair of numbers; the pair of numbers is the point. They are the same thing seen two ways.
Once that bridge is built, three translations follow at once:
• A point is a pair of numbers (x, y).
• A figure (a line, a circle, a curve) is the set of all points whose addresses satisfy one equation.
• A geometric question ("are these perpendicular? where do they cross? how far?") becomes an algebra problem you can solve by computing.
That last line is the engine of the entire stage. "Is this triangle right-angled?" used to need a careful diagram and a theorem; now it needs three subtractions and a comparison. "Where does the road meet the river?" becomes "solve two equations together." Geometry stops being a gallery of figures to admire and becomes a calculator you can run. The rest of this lesson builds the three most basic computations — distance, midpoint, and the section point — and shows how last stage's vectors quietly come back to the page.
Take two addresses, P₁(x₁, y₁) and P₂(x₂, y₂). To get from one to the other, you can always walk in two straight legs: first sideways, then up. The sideways leg has length |x₂ − x₁| (the change in x); the upward leg has length |y₂ − y₁| (the change in y). Those two legs meet at a right angle — and the straight span from P₁ to P₂ is the hypotenuse of that right triangle.
That is exactly the setup for the Pythagorean theorem (Stage 15): the square on the hypotenuse equals the sum of the squares on the legs. So
|P₁P₂| = √( (x₂ − x₁)2 + (y₂ − y₁)2 )
The straight-line distance is the square root of (the run squared plus the rise squared). It does not matter which point you call first — the differences get squared, so the sign washes out.
From A(1, 1) to B(4, 5): the run is x₂ − x₁ = 4 − 1 = 3, the rise is y₂ − y₁ = 5 − 1 = 4. So
|AB| = √(32 + 42) = √(9 + 16) = √25 = 5.
A clean 3-4-5 right triangle — the hero figure at the top shows exactly this.
The distance is √(Δx2 + Δy2), not |Δx| + |Δy|. Adding the two legs (3 + 4 = 7) measures the city-block walk along the grid — the long way round corners. The straight span cuts across and is always shorter: here 5, not 7. You must square, add, then take the root.
Now the question "where is the middle of the segment P₁P₂?" The middle point is exactly halfway across and exactly halfway up, so each coordinate is just the average of the two:
M = ( x₁ + x₂2 , y₁ + y₂2 )
Average the x's, average the y's. For A(1, 1) and B(4, 5) the midpoint is M( (1+4)/2 , (1+5)/2 ) = (2.5, 3).
The midpoint is the special case of a more flexible idea. Suppose you do not want the exact middle, but a point P that splits the segment from A to B in the ratio AP : PB = λ : 1 (read λ as "lambda"). Pushing P toward B means a bigger λ; pulling it toward A means a smaller one. The coordinates of that section point are a weighted average:
P = A + λB1 + λ ⟶ P = ( x₁ + λx₂1 + λ , y₁ + λy₂1 + λ ) for AP : PB = λ
When λ = 1 the two weights match and this collapses to the ordinary midpoint (A + B)/2 — the midpoint is "split it 1 : 1."
A surprising number of slips come from confusing the two formulas. Distance uses the difference of the coordinates (then squares). The midpoint uses the sum of the coordinates (then halves). If you ever write a minus sign inside the midpoint, stop — it should be a plus.
Last stage you carried vectors out into three dimensions. Flatten one back onto the page and it becomes a plain pair (a, b) again — and that pair has a beautiful double life. The arrow from P₁(x₁, y₁) to P₂(x₂, y₂), written P₁P₂, is exactly the displacement you walk: so many across, so many up.
P₁P₂ = (x₂ − x₁, y₂ − y₁)
Subtract tail from head, coordinate by coordinate. And its length is precisely the distance from §30.1.2: |P₁P₂| = √((x₂−x₁)2 + (y₂−y₁)2). The displacement vector and the distance are the same fact, told as an arrow instead of a number. (See vector coordinates.)
This is more than a tidy coincidence. That arrow carries a direction as well as a length — and a direction, as the next lesson shows, is exactly what fixes the slope of a line. The pair (x₂ − x₁, y₂ − y₁) is the seed of "rise over run." So the humble subtraction we just did to measure distance is quietly the same subtraction that will pin down every straight line in §30.2 · Slope and the equations of a line.
Four small computations, all built from one idea — a point is its address (x, y):
| What you want | Formula | It is… |
|---|---|---|
| Distance |P₁P₂| | √( (x₂−x₁)2 + (y₂−y₁)2 ) | the Pythagorean hypotenuse |
| Midpoint M | ( (x₁+x₂)/2 , (y₁+y₂)/2 ) | the average of the addresses |
| Section point P (AP:PB = λ) | (A + λB)/(1 + λ) | a weighted average (λ=1 → midpoint) |
| Displacement P₁P₂ | (x₂−x₁, y₂−y₁) | the arrow; its length is the distance |
Distance subtracts then squares; the midpoint adds then halves. Hold those two apart and the rest of analytic geometry follows.
Six questions to lock it in. Tap the answer you think is right.
This lesson opens Stage 30 (Analytic Geometry) by establishing the coordinate method itself: a point as an ordered pair, and the three foundational computations — distance, midpoint, and the section (partition) point. It directly serves HSG-GPE.B.7 (use coordinates to compute perimeters and areas via the distance formula), HSG-GPE.B.6 (find the point that partitions a segment in a given ratio), and HSG-GPE.B.4 (use coordinates to prove simple geometric facts, e.g. that a triangle is right-angled). The distance formula is the 8.G.B.8 / HSG-SRT Pythagorean theorem applied on the grid, and the displacement-vector view connects to HSN-VM.A from the prior stage. The supporting algebra — evaluating √ of a sum of squares and forming weighted averages — exercises HSA-REI reasoning. Encourage students to reason out loud: state the formula, show the subtraction or the average, then write the conclusion. The two interactive figures compute every length and point from the formulas (never read off the picture), so they reinforce that the diagram and the algebra must always agree.