Coordinate Geometry

Why This Matters

Imagine a town B. It is 36 km east of town A and 15 km north of it. How far apart are the two towns if you go in a straight line? Could you find this without dragging a measuring tape across the fields? Here is another one. A phone company wants to put a tower on the road from A to B. The tower must sit twice as far from B as from A. Where exactly on the map does it go?

These are not just puzzles on paper. Think about how a map works. Every place gets a pair of numbers (x, y). Once you do that, questions about distance (how far apart) and position (where something is) turn into simple arithmetic. That is the whole power of coordinate geometry. It lets you study shapes and points using algebra. You don’t need a protractor or a ruler. You just need the coordinates and a formula.

In this chapter you’ll build two tools. The first is the distance formula. It tells you how far apart two points are, using only their coordinates. The second is the section formula. It tells you the coordinates of a point that splits a line segment in a given ratio. We won’t just hand these formulas to you. We will prove both of them, starting from geometry you already know. With these two tools you can check whether three points lie on one line, whether four points form a square, and where exactly a point cuts a segment.

The Big Idea

Once you give each point coordinates (x, y), geometry becomes algebra. To find the distance between two points, draw a right triangle. Its two short sides are the gap in the x-values and the gap in the y-values. The distance is the longest side (the hypotenuse). So d = √[(x₂ − x₁)² + (y₂ − y₁)²], straight from the Pythagoras theorem. For the section formula, take a point P that splits the segment from A to B in the ratio m : n. Its coordinates are a kind of weighted average of A’s and B’s: ((mx₂ + nx₁)/(m + n), (my₂ + ny₁)/(m + n)). The midpoint is just the special case where m = n. Then it is the plain average.

Let’s Break It Down

The coordinate plane — where every point gets a name

Everything in this chapter sits on the coordinate plane. This is just two number lines that cross each other at right angles. The horizontal one is the x-axis. The vertical one is the y-axis. They meet at a point called the origin O, which is the point (0, 0). Now every point gets a name: an ordered pair (x, y). The first number tells you how far to go across (left or right along x). The second number tells you how far to go up or down (along y). For example, to plot the point (3, 2), you go 3 steps right, then 2 steps up.

The two axes cut the plane into four parts called quadrants. We number them 1, 2, 3, 4, going anticlockwise and starting from the top-right. Each quadrant has its own pattern of plus and minus signs. This is a handy way to catch mistakes. If your answer puts a point in the wrong quadrant, you know something has gone wrong. Figure 7.1 below shows the axes, the four quadrants with their sign patterns, and how the point P(3, 2) is plotted.

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Distances along an axis — the easy warm-up

Before we get to the full formula, let’s look at the easiest case. Suppose two points both lie on the x-axis, like A(4, 0) and B(6, 0). The distance between them is just the gap in their x-values: AB = 6 − 4 = 2 units. The same idea works on the y-axis. Take C(0, 3) and D(0, 8). The distance is CD = 8 − 3 = 5 units.

So when two points are on the same axis, distance is simply a subtraction. The real question is what to do when the two points are sitting anywhere in the plane, not on the same line. That is where the Pythagoras theorem helps us.

Deriving the distance formula

The whole formula leans on one rule you’ve met before. Let’s quickly refresh it before we put it to work.

Take any two points P(x₁, y₁) and Q(x₂, y₂). Draw a straight line down from P to the x-axis, and another straight line down from Q to the x-axis. Now draw a flat horizontal line starting at P. Extend it until it hits the vertical line coming down from Q. Call that meeting point T. Look at the triangle PTQ. The corner at T is a perfect right angle (90°), so PTQ is a right-angled triangle. Figure 7.2 below shows this triangle, with the two legs and the hypotenuse d labelled.

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The two short sides (the legs) of this triangle are easy to read off the picture:

• The flat side PT = x₂ − x₁ (the gap between the x-coordinates).
• The upright side QT = y₂ − y₁ (the gap between the y-coordinates).

The Pythagoras theorem says: in a right triangle, (longest side)² = (one leg)² + (other leg)². Here PQ is the longest side. So:

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

A distance can never be negative, so we take only the positive square root:

PQ = √[(x₂ − x₁)² + (y₂ − y₁)²]

That is the distance formula. Two things worth noticing:

• Each gap gets squared. So it does not matter which point you call point 1 and which you call point 2. For example, (x₂ − x₁)² = (x₁ − x₂)², because squaring a negative number makes it positive. You can subtract in either order and get the same answer.
• What if one point is the origin O(0, 0)? Then x₁ = 0 and y₁ = 0, and the formula becomes the distance of P(x, y) from the origin: OP = √(x² + y²).

Let’s see the formula in action on two simple points before anything trickier.

Now let’s settle the town puzzle we opened the chapter with — the same formula does the job.

Using distances: collinearity, triangles and quadrilaterals

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

Collinear points. “Collinear” simply means lying on one straight line. Three points A, B, C are collinear exactly when the longest of the three distances equals the sum of the other two. For example, AB + BC = AC. Why does this work? If the three points made a real triangle, then any two sides would always add up to more than the third side. So when two of the distances add up to exactly the third, the triangle has flattened out into a straight line. Figure 7.3 below shows both cases side by side: points on a line, and points forming a triangle.

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Let’s try this distance test on three actual points and see if they line up.

Types of triangle. Find all three side lengths. If all three are equal, the triangle is equilateral. If exactly two are equal, it is isosceles. And there is one more check: if the squares of two sides add up to the square of the third side, then those two sides meet at a right angle. This is the converse of the Pythagoras theorem (the theorem read backwards). We saw earlier that a 90° corner forces leg² + leg² = hypotenuse². The converse says the reverse is also true: if you find that two sides squared add up to the third side squared, the corner between them must be 90°. No other angle can make those numbers fit. So the equation itself is proof of a right angle.

Types of quadrilateral. A quadrilateral is any four-sided shape. Take the four points in order around the shape. Find the four side lengths and the two diagonals (the lines joining opposite corners). Then use this table to name the shape:

You might wonder why the diagonals matter at all. If a shape already has all four sides equal, isn’t that enough to call it a square? No — and this is exactly where students lose marks. A rhombus also has all four sides equal. Picture a square that you push sideways so it leans over, like a squashed cardboard box. The four sides stay the same length, but the corners are no longer square (90°). That tilted shape is a rhombus, not a square. So “four equal sides” cannot, on its own, tell a square apart from a rhombus.

The diagonals are what break the tie. In a square the two diagonals come out equal. In a leaning rhombus one diagonal stretches long while the other shrinks short, so they are unequal. (In fact, the diagonals come out equal only when the corners are true 90° corners — so equal diagonals are a secret way of checking for square corners without measuring any angle.) That is why the square test needs both equal sides and equal diagonals. Figure 7.4 below shows the two shapes side by side.

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Let’s put the table to work on four points and prove what shape they make.

Deriving the section formula

Now for the second tool. Suppose a point P(x, y) sits on the segment from A(x₁, y₁) to B(x₂, y₂). It divides the segment in the ratio m : n. “Divides internally in the ratio m : n” just means the part AP and the part PB are in the ratio m : n. In short, AP : PB = m : n. For example, a ratio of 3 : 1 means AP is three times as long as PB.

Now draw straight lines down from A, P and B to the x-axis. Then draw a flat horizontal line from A and another from P. This makes two right triangles, AQP and PCB. These two triangles have the same set of angles, because the segment AB crosses both of them at the same slant. When two triangles have the same angles, they have the same shape (one is just a bigger or smaller copy of the other). We call such triangles similar. This is the AA similarity criterion (AA means “angle-angle”). So △AQP ~ △PCB.

Similar triangles are the engine of this proof, so let’s make sure that idea is solid first.

Figure 7.5 below shows those two similar triangles drawn under the segment, so you can see exactly which sides we’ll compare.

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In similar triangles, matching sides are in the same ratio. That ratio here is AP : PB = m : n. So:

AQ / PC = PQ / BC = m / n

Read these sides off the figure:

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

Let’s do the x-part first. We have (x − x₁) / (x₂ − x) = m / n. Cross-multiply (multiply each side’s top by the other side’s bottom):

n(x − x₁) = m(x₂ − x)

Open the brackets: nx − nx₁ = mx₂ − mx. Bring the x-terms to one side: mx + nx = mx₂ + nx₁. Take x common on the left: (m + n)x = mx₂ + nx₁. So:

x = (mx₂ + nx₁) / (m + n)

The y-part works in exactly the same way. From (y − y₁) / (y₂ − y) = m / n we get y = (my₂ + ny₁) / (m + n).

So the point P that divides AB in the ratio m : n is:

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

This is the section formula. Watch the “cross” pattern here. The first ratio number m gets multiplied by the far point B (its x₂ and y₂). The second ratio number n gets multiplied by the near point A (its x₁ and y₁). The numbers cross over to the opposite point.

The midpoint — just the m = n case

The midpoint is the exact middle of a segment. It splits the segment into two equal halves, so the ratio is 1 : 1. Put m = 1 and n = 1 into 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: to find the midpoint, just take the average of the two x-values and the average of the two y-values. There is no new formula to memorise here. It comes straight out of the section formula.

Let’s run the section formula on a real 3 : 1 split and watch the crossed pattern do its work.

We can also run the formula backwards: given the dividing point, find the unknown ratio.

Before moving on, here’s a quick one to make sure the midpoint-as-special-case idea stuck.

Common Mistakes

These two formulas trip up almost everyone in the same few spots. Here are the slip-ups to watch for so you don’t lose easy marks.

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 used coordinates to describe positions and distances. Next, in Introduction to Trigonometry, you’ll link the angles of a right triangle to the ratios of its sides. These ratios are called sine, cosine and tangent. Think of it this way: coordinate geometry tells you how far, while trigonometry tells you which direction. Put together, they let you do amazing things, from finding the height of a tall tower to navigation and physics.