1. Descartes' bridge: geometry as algebra
Before 1637, geometry and algebra were largely separate worlds. Geometry was about diagrams — straight-edge, compass, clever construction. Algebra was about manipulating symbols. René Descartes' La Géométrie (and, independently, Pierre de Fermat) glued the two together with a single move: pick a point in the plane, call it the origin, draw two perpendicular lines through it, and use them to address every other point with a pair of numbers.
With that one trick, every geometric statement gains an algebraic translation:
- "This point lies on this line" becomes "this pair of numbers satisfies this equation."
- "These two lines are parallel" becomes "these two equations have the same slope."
- "This circle has radius $3$" becomes "$(x - h)^2 + (y - k)^2 = 9$."
Questions that took Euclid clever auxiliary lines now take a few lines of algebra. And the bridge runs both ways: every equation in two variables draws a curve. Coordinate geometry is what made calculus possible a generation later — derivatives are slopes, integrals are areas, and both need points to have numerical coordinates before you can do anything with them.
"I shall have given the complete solution of the problem if I can express all the lines of the figure by means of two unknown quantities, which I shall set down by the letters $x$ and $y$." — Descartes, La Géométrie
2. The coordinate plane
Two perpendicular number lines — the $x$-axis (horizontal) and the $y$-axis (vertical) — crossing at the origin $(0, 0)$. Every point $P$ in the plane is named by a unique ordered pair $(x, y)$: $x$ is the signed horizontal distance from the origin, $y$ the signed vertical distance.
The two axes carve the plane into four quadrants, numbered counterclockwise starting at the upper-right:
| Quadrant | Sign of $x$ | Sign of $y$ | Example |
|---|---|---|---|
| I | $+$ | $+$ | $(3, 5)$ |
| II | $-$ | $+$ | $(-3, 5)$ |
| III | $-$ | $-$ | $(-3, -5)$ |
| IV | $+$ | $-$ | $(3, -5)$ |
Points on the axes themselves belong to no quadrant: $(4, 0)$ sits on the $x$-axis, $(0, -2)$ on the $y$-axis.
An ordered pair is always $(x, y)$ — horizontal first, vertical second. Mathematicians fight wars over many notational choices, but this one is essentially universal.
3. Distance between two points
Given two points $P_1 = (x_1, y_1)$ and $P_2 = (x_2, y_2)$, how far apart are they? Draw a right triangle whose hypotenuse joins them, with legs parallel to the axes. The horizontal leg has length $|x_2 - x_1|$; the vertical leg has length $|y_2 - y_1|$. By the Pythagorean theorem:
$$ d^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2 $$Taking the positive square root gives the distance formula:
$$ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} $$The order of subtraction doesn't matter — squaring kills the sign. The formula is symmetric in the two points: $d(P_1, P_2) = d(P_2, P_1)$, which is reassuring because distance ought to be.
The distance formula isn't a new fact to memorize. It's the Pythagorean theorem written in coordinate language. Anytime you forget it, draw the triangle.
4. Midpoint and the section formula
The midpoint of the segment from $(x_1, y_1)$ to $(x_2, y_2)$ is the point exactly halfway along it. Its coordinates are the averages of the endpoints' coordinates:
$$ M = \left( \frac{x_1 + x_2}{2},\ \frac{y_1 + y_2}{2} \right) $$Intuitively, "halfway across" in $x$ and "halfway up" in $y$. There's nothing geometric to prove — averaging the coordinates lands you exactly at the centre of the rectangle whose opposite corners are $P_1$ and $P_2$.
The same idea generalises. Suppose you want the point that divides $\overline{P_1 P_2}$ in the ratio $m : n$ — that is, the point $m / (m+n)$ of the way from $P_1$ to $P_2$. The section formula gives
$$ \left( \frac{m x_2 + n x_1}{m + n},\ \frac{m y_2 + n y_1}{m + n} \right). $$The midpoint is the case $m = n = 1$. The endpoints of thirds use $m : n = 1 : 2$ and $2 : 1$.
5. Slope between two points
The slope of the line through $(x_1, y_1)$ and $(x_2, y_2)$ measures how steeply it tilts — how much $y$ changes per unit change in $x$:
$$ m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{\Delta y}{\Delta x} \quad \text{("rise over run")} $$A few cases:
| Slope | Picture | Example |
|---|---|---|
| $m > 0$ | Rises left-to-right | $(0, 0)$ to $(2, 6)$: $m = 3$ |
| $m < 0$ | Falls left-to-right | $(0, 4)$ to $(2, 0)$: $m = -2$ |
| $m = 0$ | Horizontal | $(3, 5)$ to $(7, 5)$ |
| undefined | Vertical ($\Delta x = 0$) | $(4, 2)$ to $(4, 9)$ |
Slope captures two geometric ideas at once: parallelism and perpendicularity. Two non-vertical lines are parallel iff their slopes are equal, and perpendicular iff their slopes multiply to $-1$. (Vertical and horizontal lines are perpendicular too — they're the edge case where the product rule breaks down.)
6. Equations of a line
A line in the plane can be written several ways. Each form makes a different fact about the line easy to read off; choose the one that fits what you know.
| Form | Equation | Reads off | Best when |
|---|---|---|---|
| Slope-intercept | $y = mx + b$ | Slope $m$, $y$-intercept $b$ | Graphing, comparing steepness |
| Point-slope | $y - y_1 = m(x - x_1)$ | A point and the slope | You know a point and a slope |
| Standard | $Ax + By = C$ | Symmetric in $x, y$; integer coefficients | Solving systems, clean algebra |
| General | $Ax + By + C = 0$ | Implicit; treats $x$ and $y$ alike | Distance from point to line, normals |
All four are interchangeable — rearranging is just algebra. For example, $2x + 3y = 12$ (standard) becomes $y = -\tfrac{2}{3}x + 4$ (slope-intercept) by solving for $y$, or $2x + 3y - 12 = 0$ (general) by moving everything to one side. The line they describe is the same; the form just controls what's foregrounded.
For a complete tour of the slope-intercept story — sliders, intercepts, the works — see Linear Equations in the Algebra chapter. Here we'll lean on the general form because the next section needs it.
A vertical line has no slope, so it has no slope-intercept form. It's written $x = c$ for some constant $c$ — every point on it has the same $x$-coordinate, and $y$ is free to be anything.
7. Distance from a point to a line
Given a point $P_0 = (x_0, y_0)$ and a line $\ell$, the distance from $P_0$ to $\ell$ is the length of the perpendicular segment from $P_0$ to $\ell$ — the shortest possible bridge between them. Write the line in general form $ax + by + c = 0$. Then:
$$ d(P_0, \ell) = \frac{|a x_0 + b y_0 + c|}{\sqrt{a^2 + b^2}} $$Two things to notice. First, $a x_0 + b y_0 + c$ is exactly what you'd get by plugging the point into the left side of the equation: if the point is on the line, that quantity is zero, so the distance is zero. Good. Second, the absolute value is essential — without it, the formula gives a signed distance whose sign tells you which side of the line $P_0$ sits on. Both versions are useful; just know which one you want.
A handy consequence: two parallel lines $ax + by + c_1 = 0$ and $ax + by + c_2 = 0$ are separated by
$$ \frac{|c_2 - c_1|}{\sqrt{a^2 + b^2}}, $$because every point on one line is the same distance from the other.
The formula only works when the line is in the form $ax + by + c = 0$ — everything on one side, zero on the other. If you're handed $y = 2x + 3$, rewrite it as $2x - y + 3 = 0$ before plugging in. A common mistake is to use $y - (2x + 3)$ as if that's already in standard form; it leads to the wrong sign on $a$ and a numerically correct but conceptually muddled answer.
8. Equations of circles
A circle is the set of points at a fixed distance $r$ (the radius) from a fixed point $(h, k)$ (the centre). Translate that definition through the distance formula and the algebraic form drops out:
$$ \sqrt{(x - h)^2 + (y - k)^2} = r $$Square both sides to clear the root:
$$ (x - h)^2 + (y - k)^2 = r^2 $$This is the standard form of a circle. A point $(x, y)$ lies on the circle iff its coordinates make the equation true; the equation is the circle.
Common cases:
- Centre at the origin, radius $r$: $x^2 + y^2 = r^2$.
- Unit circle (origin, radius $1$): $x^2 + y^2 = 1$.
- Centre $(2, -3)$, radius $5$: $(x - 2)^2 + (y + 3)^2 = 25$.
If you multiply out the squares you get the general form $x^2 + y^2 + Dx + Ey + F = 0$. Going the other way — recovering centre and radius from general form — is done by completing the square in $x$ and $y$ separately.
"Set of points at distance $r$ from $(h, k)$" is a purely geometric description. "$(x - h)^2 + (y - k)^2 = r^2$" is a purely algebraic one. They are the same object. Every question about the circle — does this line intersect it? Where? What's the tangent at this point? — can now be answered with algebra alone.
9. The geometry–algebra dictionary
Once you have coordinates, every geometric statement has an algebraic translation. The fluent move is to glance at a problem, decide which side of the dictionary is easier, and work there. The table below is a starter:
| Geometric statement | Algebraic translation |
|---|---|
| Point $P$ lies on line $\ell$ | $(x_P, y_P)$ satisfies the equation of $\ell$ |
| Two lines are parallel | Equal slopes (or both vertical) |
| Two lines are perpendicular | Slope product $= -1$ (or one vertical, one horizontal) |
| Three points are collinear | Slopes between pairs agree; equivalently, triangle area $= 0$ |
| Lines $\ell_1, \ell_2$ intersect | The system of their equations has a solution |
| Point lies inside circle of radius $r$ around $(h, k)$ | $(x - h)^2 + (y - k)^2 < r^2$ |
| Two figures are congruent (translated) | One equation becomes the other under $x \mapsto x + a$, $y \mapsto y + b$ |
The dictionary's true power is that algebra is mechanical. You don't need to be clever; you need to plug in, simplify, and read off. That is exactly what made Descartes' move revolutionary, and it's what makes coordinate geometry the workhorse of physics, engineering, computer graphics, and machine learning.
"And as I considered the matter carefully it gradually came to light that all those matters only were referred to mathematics in which order and measurement are investigated." — Descartes, Rules for the Direction of the Mind
10. Common pitfalls
The slope of a vertical line is undefined, not zero. Zero is horizontal. If your slope formula produces "divide by zero," you've hit a vertical line — switch to $x = c$ rather than $y = mx + b$.
$m = (y_2 - y_1) / (x_2 - x_1)$. Swapping the two points flips both the numerator and the denominator, so the sign is preserved — but doing one swap and not the other gives the wrong sign. Subtract in the same order, both times.
You cannot read slope and intercept directly off $2x - 3y = 6$. You must first rearrange to $y = \tfrac{2}{3}x - 2$. The coefficient of $x$ in the original equation is not the slope.
The point-to-line distance formula has $|ax_0 + by_0 + c|$ in the numerator. Without the absolute value you get a signed quantity that can be negative — useful for "which side" questions, but never a distance.
In $(x - h)^2 + (y - k)^2 = r^2$ the centre is $(h, k)$ — the values you subtract, not what's written inside the parentheses. So $(x + 2)^2 + (y - 3)^2 = 25$ has centre $(-2, 3)$, not $(2, -3)$, and radius $5$, not $25$. Read carefully.
11. Worked examples
Try each one before opening the solution.
Example 1 · Distance from $(1, 2)$ to $(4, 6)$
Step 1. Compute the coordinate differences:
$$ \Delta x = 4 - 1 = 3, \qquad \Delta y = 6 - 2 = 4 $$Step 2. Apply the distance formula:
$$ d = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5 $$A classic 3-4-5 right triangle in disguise.
Example 2 · Midpoint of the segment from $(2, 8)$ to $(10, 2)$
Average each coordinate:
$$ M = \left( \frac{2 + 10}{2},\ \frac{8 + 2}{2} \right) = (6, 5) $$That's it. The midpoint is the point with the average address.
Example 3 · Equation of the line through $(1, 2)$ and $(4, 11)$
Step 1. Compute the slope:
$$ m = \frac{11 - 2}{4 - 1} = \frac{9}{3} = 3 $$Step 2. Use point-slope form with $(1, 2)$:
$$ y - 2 = 3(x - 1) $$Step 3. Simplify to slope-intercept:
$$ y = 3x - 3 + 2 = 3x - 1 $$Check. At $x = 4$: $y = 3(4) - 1 = 11$. ✓
Example 4 · Distance from $(3, 4)$ to the line $3x + 4y - 5 = 0$
Identify $a = 3$, $b = 4$, $c = -5$, $(x_0, y_0) = (3, 4)$. Plug in:
$$ d = \frac{|3(3) + 4(4) - 5|}{\sqrt{3^2 + 4^2}} = \frac{|9 + 16 - 5|}{\sqrt{25}} = \frac{20}{5} = 4 $$The point sits at distance $4$ from the line.
Example 5 · Find the centre and radius of $x^2 + y^2 - 4x + 6y - 12 = 0$
Step 1. Group $x$ and $y$ terms:
$$ (x^2 - 4x) + (y^2 + 6y) = 12 $$Step 2. Complete the square in each group. For $x^2 - 4x$, add $4$ to make $(x - 2)^2$. For $y^2 + 6y$, add $9$ to make $(y + 3)^2$. Add the same constants to the right side:
$$ (x - 2)^2 + (y + 3)^2 = 12 + 4 + 9 = 25 $$Step 3. Read off centre $(2, -3)$ and radius $\sqrt{25} = 5$.
Example 6 · Are $(1, 2)$, $(3, 6)$, $(4, 8)$ collinear?
Compute the slope between the first two points and the slope between the last two:
$$ m_{12} = \frac{6 - 2}{3 - 1} = 2, \qquad m_{23} = \frac{8 - 6}{4 - 3} = 2 $$The slopes agree, so the three points lie on the same line. Collinear.