Analytic Geometry
The Cartesian Plane, Distance and Midpoint
Ordered pairs, the distance formula as Pythagoras in disguise, and the midpoint as an average of coordinates.
What this page covers
- Plot and read an ordered pair on the coordinate plane
- Derive and apply the distance formula
- Find the midpoint of a segment
- Explain why the order of the two points does not matter
The plane and the ordered pair
Analytic geometry is the idea that a geometric point can be named by a pair of numbers, so that geometric questions become algebraic ones. The source attributes the idea and then gets straight to the notation: a point is an ordered pair (x, y).
Ordered is the operative word. (3, 7) and (7, 3) are different points. The first coordinate is measured along the horizontal axis, the second along the vertical.
- x-coordinate (abscissa)
- Horizontal position, positive to the right
- y-coordinate (ordinate)
- Vertical position, positive upward
- Origin
- (0, 0), where the axes meet
- Quadrants
- I: both positive. II: x < 0, y > 0. III: both negative. IV: x > 0, y < 0
The distance formula
Two points and the horizontal and vertical lines joining them form a right triangle. The legs have lengths |x2 - x1| and |y2 - y1|, and the distance sought is the hypotenuse.
The absolute values disappear on squaring, which is why the formula needs none. It also means the order of the two points is irrelevant — a point the source makes by computing an example twice.
Worked example — the source's own
Find the distance between (-3, 7) and (1, -4).
Taking the points in the other order:
1 - (-3) = 4, not -2. Two of the four subtractions here involve a negative coordinate, and both are places where a sign is commonly lost. Writing the substitution with explicit brackets prevents it.
√137 ≈ 11.7. The vertical separation alone is 11, and the hypotenuse must exceed either leg but be less than their sum (4 + 11 = 15). The answer sits correctly between.
The midpoint
Each coordinate is the average of the corresponding coordinates of the endpoints, because moving halfway along the segment moves halfway in each direction independently.
Worked example — the source's case
Find the midpoint of the segment joining (-3, 7) and (4, -2).
The midpoint should be the same distance from each endpoint. From (-3, 7): √(3.5)2 + (4.5)2 = √32.5. From (4, -2): √(3.5)2 + (4.5)2 = √32.5. Equal.
Add the coordinates, do not subtract them. The distance formula uses differences and the midpoint formula uses sums, and mixing the two is a frequent slip when both appear in the same question.
What the two formulae are used for
| Task | Uses | How |
|---|---|---|
| Equation of a circle | Distance | The circle is the set of points a fixed distance from the centre |
| Testing for an isosceles triangle | Distance | Compute all three side lengths and compare |
| Testing for a right angle | Distance | Check whether a2 + b2 = c2 |
| Centre of a circle from a diameter | Midpoint | The centre is the midpoint of any diameter |
| Perpendicular bisector | Both | Passes through the midpoint, perpendicular to the segment |
| Magnitude of a vector | Distance | |v| = √v12 + v22 is the same formula |
The last row is worth noting. The magnitude of a vector and the distance between two points are the same computation; a vector's components are exactly the coordinate differences of its endpoints, as the source observes in Week 10.
Common mistakes
| Mistake | Correct | Check |
|---|---|---|
| 1 - (-3) = -2 | = 4 | Bracket every substitution |
| Averaging for distance | Distance uses differences | Distance is never zero for distinct points |
| Subtracting for the midpoint | Midpoint uses sums | The midpoint must lie between the endpoints |
| √a2 + b2 = a + b | No such rule | √16 + 121 = 11.7, not 15 |
| Forgetting to square before adding | Square each difference first | Order of operations inside the radical |
Frequently asked questions
Does it matter which point I call the first?
No. The differences are squared, so (x2 - x1)2 equals (x1 - x2)2. The source computes one example both ways and gets √137 each time.
Why is the midpoint just an average?
Because the midpoint is halfway along in each coordinate independently. Halfway between x1 and x2 is their mean, and the same for y.
Is the distance formula a separate result?
No. It is Pythagoras applied to the right triangle whose legs are the horizontal and vertical separations. Nothing new is being asserted.
Does it extend to three dimensions?
Yes. d = √(x2 - x1)2 + (y2 - y1)2 + (z2 - z1)2, by applying Pythagoras twice.
Source. Handwritten teaching notes, Week 4, pages 6-7 and 12.
This page is an original exposition prepared for the KEVOS® knowledge library. It restates, reorganises and verifies the mathematics of the supplied teaching notes; it is not a reproduction of them. Numerical values taken from the notes are identified as source examples and are not presented as engineering standards.