Coordinate Geometry

Why This Matters

Imagine a town B that lies 36 km east and 15 km north of town A. How far apart are the two towns — in a straight line — without ever stretching a measuring tape across the countryside? Or: a phone company wants to place a relay tower on the road from A to B so that it sits twice as far from B as from A — exactly where on the map does it go?

These aren’t drawing-board puzzles. The moment you pin every point on a map to a pair of numbers (x, y), questions about distance and position become pure arithmetic. That is the whole power of coordinate geometry: it lets you study geometry using algebra. No protractor, no ruler — just the coordinates and a formula.

In this chapter you’ll build two tools that do exactly this. The distance formula tells you how far apart any two points are from their coordinates alone. The section formula tells you the coordinates of a point that splits a segment in a given ratio. Both come straight out of the geometry you already know — and we’ll prove them, not just hand them to you. With these two tools you can decide whether three points lie on a line, whether four points form a square, and where exactly a point divides a segment.

The Big Idea

Once each point is labelled with coordinates (x, y), geometry turns into algebra. The distance between two points is just the hypotenuse of a right triangle whose legs are the differences in their x- and y-coordinates — so d = √[(x₂ − x₁)² + (y₂ − y₁)²] straight from the Pythagoras theorem. And a point P that divides the segment from A to B in the ratio m : n has coordinates that are a weighted average of A’s and B’s — ((mx₂ + nx₁)/(m + n), (my₂ + ny₁)/(m + n)). The midpoint is simply the m = n case: the plain average.

Let’s Break It Down

Distances along an axis — the easy warm-up

Before the general formula, look at the simplest case. If two points lie on the x-axis, say A(4, 0) and B(6, 0), the distance is just the gap in their x-values: AB = 6 − 4 = 2 units. Likewise two points on the y-axis, say C(0, 3) and D(0, 8), are CD = 8 − 3 = 5 units apart.

So along a single axis, distance is just a subtraction. The interesting question is two points sitting anywhere in the plane — and that’s where Pythagoras comes in.

Deriving the distance formula

Take any two points P(x₁, y₁) and Q(x₂, y₂). Drop a perpendicular from P and from Q to the x-axis, and draw a horizontal line from P that meets the vertical through Q at a point T. Now PTQ is a right-angled triangle, with the right angle at T.

Tap to zoom

Read the two legs straight off the picture:

• The horizontal leg PT = x₂ − x₁ (the difference in x-coordinates).
• The vertical leg QT = y₂ − y₁ (the difference in y-coordinates).

Now apply the Pythagoras theorem to the right triangle PTQ:

PQ² = PT² + QT² = (x₂ − x₁)² + (y₂ − y₁)²

Since a distance is never negative, take only the positive square root:

PQ = √[(x₂ − x₁)² + (y₂ − y₁)²]

That is the distance formula. Two useful remarks:

• Because the differences get squared, it doesn’t matter which point you call “1” and which “2” — (x₂ − x₁)² = (x₁ − x₂)². You can subtract in either order.
• Putting one point at the origin O(0, 0) gives the distance of P(x, y) from the origin: OP = √(x² + y²).

Using distances: collinearity, triangles and quadrilaterals

Once you can find any distance, you can answer “what shape is this?” purely by comparing lengths.

Collinear points. Three points A, B, C lie on a single straight line exactly when the longest of the three distances equals the sum of the other two — e.g. AB + BC = AC. (If they made a triangle, any two sides would add up to more than the third; equality means the triangle has collapsed flat onto a line.)

Types of triangle. Compute all three sides. All three equal → equilateral; exactly two equal → isosceles. And if the squares of two sides add up to the square of the third, the converse of the Pythagoras theorem says there’s a right angle there.

Types of quadrilateral. For four points (taken in order), compute the four sides and the two diagonals, then read off the shape:

Deriving the section formula

Now the second tool. Suppose a point P(x, y) lies on the segment from A(x₁, y₁) to B(x₂, y₂) and divides it internally in the ratio m : n — meaning AP : PB = m : n.

Drop perpendiculars from A, P and B to the x-axis, and draw horizontal lines from A and from P. This creates two right triangles, AQP and PCB, sharing equal angles (the segment AB cuts across both at the same slope). By the AA similarity criterion, △AQP ~ △PCB.

Tap to zoom

From the similar triangles, corresponding sides are in the same ratio AP : PB = m : n, so:

AQ / PC = PQ / BC = m / n

Read those legs off the figure:

• AQ = x − x₁ and PC = x₂ − x.
• PQ = y − y₁ and BC = y₂ − y.

Take the x-part first: (x − x₁) / (x₂ − x) = m / n. Cross-multiply:

n(x − x₁) = m(x₂ − x) → nx − nx₁ = mx₂ − mx → mx + nx = mx₂ + nx₁ → x = (mx₂ + nx₁) / (m + n)

The y-part works identically: (y − y₁) / (y₂ − y) = m / n gives y = (my₂ + ny₁) / (m + n).

So the point P dividing AB internally in the ratio m : n is:

P = ((mx₂ + nx₁)/(m + n), (my₂ + ny₁)/(m + n))

This is the section formula. Notice the “cross” pattern: the x of the far point B gets multiplied by m (the first ratio number), and the x of the near point A by n.

The midpoint — just the m = n case

The midpoint divides a segment in the ratio 1 : 1, so put m = n = 1 in the section formula:

Midpoint = ((1·x₂ + 1·x₁)/(1 + 1), (1·y₂ + 1·y₁)/(1 + 1)) = ((x₁ + x₂)/2, (y₁ + y₂)/2)

In plain words: the midpoint’s coordinates are just the averages of the endpoints’ coordinates. No new formula to memorise — it falls right out of the section formula.

Common Mistakes

Quick Check

Practice Problems

Easy

Medium

Challenge

Summary

You should now be able to explain:

• Coordinate geometry turns geometric questions about points into algebra, using their coordinates (x, y).
• The distance formula: PQ = √[(x₂ − x₁)² + (y₂ − y₁)²], derived from the Pythagoras theorem on the right triangle with legs x₂ − x₁ and y₂ − y₁. The order of subtraction doesn’t matter because the differences are squared.
• The distance from the origin: OP = √(x² + y²).
• Collinearity: three points lie on a line when the largest distance equals the sum of the other two (AB + BC = AC).
• You can classify triangles and quadrilaterals by comparing side lengths and diagonals (square, rhombus, rectangle, parallelogram).
• The section formula: a point dividing AB internally in ratio m : n is ((mx₂ + nx₁)/(m + n), (my₂ + ny₁)/(m + n)) — derived from similar triangles, with the ratio numbers crossed onto the opposite endpoints.
• The midpoint formula is the m = n = 1 case: ((x₁ + x₂)/2, (y₁ + y₂)/2) — just the average of the coordinates.

What’s Next

So far you’ve described positions and distances with coordinates. Next, in Introduction to Trigonometry, you’ll connect the angles of a right triangle to the ratios of its sides — sine, cosine and tangent. Where coordinate geometry measures how far, trigonometry measures which direction, and together they power everything from measuring the height of a tower to navigation and physics.