The Timpson Lyceum · Geometry (2024)

Analytic Geometry and Equations of Lines – Lecture 18 (Geometry)

Analytic Geometry and Equations of Lines – Lecture 18

Overview

Analytic geometry fuses algebra with geometry: it coordinatizes the plane so that geometric figures can be represented by equations and studied with numbers. This approach—credited to René Descartes, which is why the framework is called the Cartesian coordinate system—replaces the case-by-case "epiphany" of synthetic proofs with systematic algebraic computation. This lecture builds the coordinate system from scratch (first a line, then a plane), establishes the one-to-one correspondence between linear equations and lines, and derives the standard forms of a line's equation. It closes with the distance formula, the midpoint formula, and the equation of a circle. A recurring theme is that although a coordinate system is entirely arbitrary (relative), the quantities it lets us compute—lengths, similarity, areas—are invariant.

Key Concepts

Coordinatizing a line. Choose a unit (a reference segment) and an arbitrary point $O$ (the origin). For any point $P_x$, its coordinate is the length of segment $\overline{P_x O}$ if $P_x$ lies on the chosen positive side of $O$, and the negative of that length if it lies on the opposite side. By convention the line is drawn horizontally with the positive side to the right. A point is a geometric object; a coordinate is a number. They are not the same thing.

Coordinatizing the plane. Take two perpendicular lines intersecting at $O$, pick a unit, and coordinatize each line as above. The horizontal line is the axis of abscissas (the $x$-axis); the vertical line is the axis of ordinates (the $y$-axis). The origin has coordinates $(0,0)$ — a consequence of the construction, not a separate definition.

Assigning coordinates to a point. To find the coordinates of a point $P$:

• Draw the line through $P$ parallel to the $y$-axis; where it meets the $x$-axis gives the abscissa $a$ (the $x$-coordinate).
• Draw the line through $P$ parallel to the $x$-axis; where it meets the $y$-axis gives the ordinate $b$ (the $y$-coordinate).
• So $P = (a,b)$ means exactly these two perpendicular projections.

Space is relative; geometric quantities are not. How you impose a coordinate system is completely arbitrary—two observers can disagree on a point's coordinates. But if they use the same unit, they agree on lengths. Distance is an invariant: it does not depend on the chosen reference frame. This is why arbitrary coordinatizing still yields absolute (non-relative) conclusions about congruence, similarity, and area.

Vertical and horizontal lines.

• $x = a$ is the unique line parallel to the $y$-axis through abscissa $a$: every point on it has $x$-coordinate $a$, and every point with $x$-coordinate $a$ lies on it.
• $y = b$ is the unique line parallel to the $x$-axis through ordinate $b$.
• A line is geometric; an equation is algebraic. The lecture's central claim is a one-to-one correspondence between them.

Linear equations and lines. An equation of the form $ax + by + c = 0, \quad \text{with } a \neq 0 \text{ or } b \neq 0,$ is called a linear equation. There is a one-to-one correspondence between such equations and lines: every linear equation describes exactly one line (the set of all $(x,y)$ satisfying it), and every line is described by exactly one such equation (up to logical equivalence). The proof splits into cases — $a=0$ gives a horizontal line $y = -c/b$; $b=0$ gives a vertical line $x = -c/a$; and the general case ($a,b \neq 0$) is proved using similar triangles (equal slopes), since slope arguments ultimately reduce to triangle similarity.

Standard forms of a line. For points $(x_1,y_1)$ and $(x_2,y_2)$ on the line:

• Two-point form: $\displaystyle y - y_1 = \frac{y_2 - y_1}{x_2 - x_1}\,(x - x_1)$
• Point-slope form: $y - y_1 = m(x - x_1)$, where $m = \dfrac{y_2 - y_1}{x_2 - x_1}$ is the slope (rise over run).
• Slope-intercept form: $y = mx + b$, where $b$ is the $y$-intercept (the value of $y$ when $x = 0$).

Distance formula. In one dimension the distance between coordinates $x_1$ and $x_2$ is $|x_2 - x_1|$ (the absolute value removes order-dependence). In two dimensions, build a right triangle with legs $|x_2 - x_1|$ and $|y_2 - y_1|$ and apply the Pythagorean theorem: $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}.$ Squaring discards the need for absolute values. The formula extends to higher dimensions by adding one squared term per coordinate (e.g. $+\,(z_2 - z_1)^2$ in three dimensions), provable by induction.

Midpoint formula. The midpoint of the segment from $(x_1,y_1)$ to $(x_2,y_2)$ is the coordinate-wise average: $\left( \frac{x_1 + x_2}{2}, \; \frac{y_1 + y_2}{2} \right).$ It is proved by noting that a line through the midpoint parallel to an axis bisects the corresponding leg of the right triangle.

Equation of a circle. A circle is the set of points at fixed distance $r$ (the radius) from a center. Applying the distance formula:

• Centered at the origin: $x^2 + y^2 = r^2$.
• Centered at $(h,k)$: $(x - h)^2 + (y - k)^2 = r^2$.

Worked Example

Find the distance between the points $(-6, 2)$ and $(-2, 5)$.

As a second illustration, find the equation of the line through $(5, -3)$ and $(-2, 4)$ using the two-point form. Here $m = \dfrac{4 - (-3)}{-2 - 5} = \dfrac{7}{-7} = -1$:

Summary

• Analytic (Cartesian) geometry coordinatizes the plane so geometric figures become equations; choose a unit and origin, then project a point onto each perpendicular axis to read off its $(x, y)$ coordinates.
• A coordinate system is arbitrary, but lengths and other geometric quantities computed in it are invariant—independent of the chosen reference frame.
• There is a one-to-one correspondence between linear equations $ax + by + c = 0$ (with $a$ or $b$ nonzero) and lines; $x = a$ and $y = b$ are the special vertical and horizontal cases.
• A line has three interchangeable forms: two-point, point-slope $y - y_1 = m(x - x_1)$, and slope-intercept $y = mx + b$, with slope $m$ as rise over run.
• The distance formula $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$ follows from the Pythagorean theorem, and the midpoint is the coordinate-wise average of the endpoints.
• A circle of radius $r$ centered at $(h, k)$ has equation $(x - h)^2 + (y - k)^2 = r^2$, derived directly from the distance formula.