The Coordinate Plane

Set theory built this room

From the Ordered Pairs and Cartesian Products lesson in the Sets stratum: the Cartesian product $\mathbb{R} \times \mathbb{R}$ is the set of all ordered pairs $(x, y)$ where $x, y \in \mathbb{R}$. That set IS the coordinate plane. Descartes didn't invent a new thing — he gave $\mathbb{R}^2$ a spatial interpretation. The $x$-axis is the first copy of $\mathbb{R}$; the $y$-axis is the second copy; the origin $(0, 0)$ is the pair $(0, 0)$.

A point is an ordered pair. $(3, -2)$ is the pair where the first component is 3 and the second is $-2$. It lives 3 units right of the origin and 2 units below it. The order matters: $(3, -2) \ne (-2, 3)$ — that's the definition of "ordered" from the Sets lesson.

Quadrants. The axes divide the plane into four quadrants:

• Quadrant I: $x > 0$, $y > 0$ (upper right)
• Quadrant II: $x < 0$, $y > 0$ (upper left)
• Quadrant III: $x < 0$, $y < 0$ (lower left)
• Quadrant IV: $x > 0$, $y < 0$ (lower right)

Points on the axes are not in any quadrant.

The distance formula: Pythagoras, translated

The Pythagorean theorem (full proof lives in the Geometry stratum) says: in a right triangle with legs $a$ and $b$ and hypotenuse $c$, we have $a^2 + b^2 = c^2$.

Now take two points $P_1 = (x_1, y_1)$ and $P_2 = (x_2, y_2)$. Draw the horizontal segment from $P_1$ to $(x_2, y_1)$ — its length is $|x_2 - x_1|$. Draw the vertical segment from $(x_2, y_1)$ to $P_2$ — its length is $|y_2 - y_1|$. These two segments are the legs of a right triangle with hypotenuse $\overline{P_1 P_2}$.

Pythagoras gives: $d(P_1, P_2)^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2.$

Taking the non-negative square root (since distance is non-negative by definition): $d(P_1, P_2) = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}.$

This is not a new formula — it's the Pythagorean theorem wearing coordinate clothes. File this away: in linear algebra, $\|v\|$ will be this formula again, in $n$ dimensions.

Example: distance from $(1, 2)$ to $(4, 6)$: $d = \sqrt{(4-1)^2 + (6-2)^2} = \sqrt{9 + 16} = \sqrt{25} = 5.$

The midpoint formula: averaging coordinates

The midpoint $M$ of the segment from $P_1 = (x_1, y_1)$ to $P_2 = (x_2, y_2)$ is halfway between them in each coordinate: $M = \left(\frac{x_1 + x_2}{2},\; \frac{y_1 + y_2}{2}\right).$

Why? Because the midpoint of any interval $[a, b]$ on the number line is the average $\frac{a+b}{2}$ — we're just applying that to each coordinate independently. The $x$ and $y$ directions are independent (they're separate copies of $\mathbb{R}$), so averaging each one separately gives the point that sits exactly halfway.

Example: midpoint of $(-3, 5)$ and $(7, 1)$: $M = \left(\frac{-3+7}{2}, \frac{5+1}{2}\right) = (2, 3).$

The graph of an equation: a set of points

The graph of an equation in $x$ and $y$ is the set of all ordered pairs $(x, y)$ that make the equation true. This is the bridge between algebra and geometry — algebra describes a condition, geometry makes it visible.

For example, the graph of $y = 2x - 1$ is the set $\{(x, y) \in \mathbb{R}^2 : y = 2x - 1\}$. When you "plot the graph" you are drawing this set. The set happens to be a straight line, but that's not obvious until you actually plot it or derive it algebraically (we'll do that in the Lines and Slope lesson).

Checking whether a point is on a graph: substitute its coordinates into the equation and see if the equation becomes true. $(3, 5)$ on $y = 2x - 1$? Substitute: $5 = 2(3) - 1 = 5$. Yes. $(0, 0)$? $0 = 2(0) - 1 = -1$. No.

The graph is a geometric object; the equation is an algebraic one; they describe the same mathematical set. This duality is the engine of the rest of this course.