Part II

Geometry & Functions

Chapter 4

Basics of Coordinate Geometry

Coordinate Geometry: The Basic Idea

Coordinate geometry is a way of thinking: Instead of thinking of a geometric figure as a complete entity, we think of it as an infinite collection of points. $ ^{*} $ This is not, of course, the way people usually think: We normally view the moon as a whole – as the moon; rarely do we think of it as an enormous collection of atoms, much less an infinite collection of mathematical points. Such, however, is the way of coordinate geometry.

Let us begin. Pick a point in the plane and draw horizontal and vertical lines through it. We call the point the origin, and the lines the coordinate axes. Relative to these axes, we can specify any point’s location with two numbers, the point’s coordinates. The figure at right shows how this works: Point P’s coordinates are $ (3, -1) $ since it lies 3 units right of and 1 unit below the relevant axes.

Next, we associate each axis with a variable.

This will let us harness our coordinates to algebra.

Unless otherwise specified, the horizontal axis will always be our x-axis, and the vertical our y-axis. $ ^{} $

A graph, in coordinate geometry, is just a fancy term for any geometric object in the plane. Lines and squiggles of all sorts, including your signature, are graphs. By thinking of graphs as collections of points, and then thinking of these points as coordinates (i.e. as pairs of numbers), we build a bridge from the tangible geometric world to the abstract numerical world, and from there, to the world of algebra.

Point | x | y First Quadrant | 1 | 2 Third Quadrant | -3 | -1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | 3 | -1 Fourth Quadrant | -1 | 1 Fourth Quadrant | -1 | 1 Fourth Quadrant | -1 | 2 Fourth Quadrant | -1 | 3 Fourth Quadrant | -1 | 4 Fourth Quadrant | -1 | 5 Fourth Quadrant | -1 | 6 Fourth Quadrant | -1 | 7 Fourth Quadrant | -1 | 8 Fourth Quadrant | -1 | 9 Fourth Quadrant | -1 | 10 Fourth Quadrant | -1 | 11 Fourth Quadrant | -1 | 12 Fourth Quadrant | -1 | 13 Fourth Quadrant | -1 | 14 Fourth Quadrant | -1 | 15 Fourth Quadrant | -1 | 16 Fourth Quadrant | -1 | 17 Fourth Quadrant | -1 | 18 Fourth Quadrant | -1 | 19 Fourth Quadrant | -1 | 20 Fourth Quadrant | -1 | 21 Fourth Quadrant | -1 | 22 Fourth Quadrant | -1 | 23 Fourth Quadrant | -1 | 24 Fourth Quadrant | -1 | 25 Fourth Quadrant | -1 | 26 Fourth Quadrant | -1 | 2

Coordinate geometry lets us describe graphs algebraically. The process requires three steps. First, we mentally atomize a graph, thinking of it as an infinite set of points. Second, we think of each point as a coordinate pair. (So after two steps, we think of our graph as an infinitely long list of pairs of numbers.) Third, we use algebra to compress our infinite list of coordinate pairs into a finite package: a single equation satisfied by the coordinates of every point on the graph – and by no others. This, the equation of the graph, is an algebraic portrait of a geometric object.

For a simple example, consider the graph at right, a line which bisects the first and third quadrants. It passes through $ (1,1) $ , $ (2,2) $ , $ (-1.4,-1.4) $ , and every other point with equal coordinates. In short, a point $ (x,y) $ in the plane lies on the line if and only if x = y. This being the case, the equation of the line is x = y.

Coordinate geometry provides algebra-tinted glasses through which we can view geometry, and geometry-tinted glasses through which we can view algebra. It is built upon a single principle, which is the basis for all applications of coordinate geometry. Understand it, and never forget it:

The Fundamental Principle of Coordinate Geometry (FPCG).

A point lies on a graph if – and only if – its coordinates satisfy the graph’s equation.

Example 1. The graph at right seems to pass through the point $ (-1,1) $ . Does it really?

Solution. Substituting -1 and 1 for x and y respectively in the graph’s equation yields

\[ (-1)^{3}+(-1\cdot1)^{2}-(-1)=1, \]

which is a true statement, as you should verify. Therefore, the point’s coordinates do satisfy the equation. Consequently, the FPCG tells us that $ (-1,1) $ does indeed lie on the graph.

A graph’s equation relates the two coordinates of every point on the graph: If we know one coordinate of such a point, then we can use the graph’s equation to find the point’s other coordinate.

Example 2. The equation of the graph at right (an ellipse) is $ 3x{2}-2xy+4y{2}+11x-6y=54 $ . Find the exact coordinates of the two points at which the ellipse crosses the x-axis.

Solution. Both points have y-coordinate 0. According to the ellipse’s equation, when y = 0, we have $ 3x^{2} + 11x = 54 $ .

Solving this yields $ x = $ .

Hence, the two points are $ (,0) $ and $ (,0) $ .

Sometimes symmetry considerations can be useful, as in this next example.

Example 3. The figure at right (a parabola) is the graph of the equation $ y = -x^{2} + 5x - 4 $ . Find the coordinates of the parabola’s peak.

Solution. Symmetry ensures that the graph’s peak is directly above the midpoint of segment AB. Thus, the peak’s x-coordinate must be 5/2 (since 5/2 is the average of 1 and 4).

When x = 5/2 for a point on the parabola, it follows from the parabola’s equation that $ y = -(5/2)^{2} + 5(5/2) - 4 = 9/4 $ .

Hence, the peak occurs at the point $ (5/2, 9/4) $ .

Exercises

    1. Does the point $ (-4,2) $ lie on the graph in Example 1? What about $ (1,1) $ ?

  1. Find another point that lies on it, and prove that it lies on it.

  2. Does the graph ever cross the y-axis (perhaps at some point light-years above the x-axis)? If so, precisely where does it cross? If not, how can you be certain that it never crosses?

  1. At which point does the parabola in Example 3 cross the y-axis?

  2. There is exactly one point on the parabola in Example 3 whose x-coordinate is 6. Find its y-coordinate.

  3. Find the exact coordinates of the two points where the ellipse in Example 2 crosses the y-axis.

  4. Two points on Example 2’s ellipse have x-coordinate 1. Locate them on the graph and find their coordinates.

Straight Lines: Preliminaries

So much for the general idea of coordinate geometry. Let’s dig into specifics, beginning with straight lines. The special cases of vertical and horizontal lines are trivial, as the next example shows.

Problem. Find the equation of the vertical line passing through $ (2, 0) $ .

Solution. By definition, the line’s equation is satisfied by the coordinates of all points on the line – and by no others. Can we think of such an equation?

The answer is obvious: Every point on the vertical line (and no point off it) has an x-coordinate of 2, so the line’s equation is simply x = 2.

By similar logic, the equation of the horizontal line that lies 3 units below the x-axis is y = -3.

The equation of an oblique line (neither vertical nor horizontal) is more involved. To find it, we must specify the feature that distinguishes straight lines from all other graphs, and translate it into algebraic language. Most curves rise more steeply at some points than others. The distinguishing feature of a line is that it is equally steep everywhere. We measure this steepness by a number called the line’s slope.

Definition. A line’s slope is the number of units it rises vertically for each unit that it runs horizontally.

For example, suppose we know that the slope of the line in the figure at right is 2.3. Between points A and B, this line runs horizontally by 5 units. As it does so, by how many units does it rise vertically? Since we know the line’s slope, this is no mystery at all (despite the figure’s question mark); the line must rise vertically by precisely $ 2.3(5) = 11.5 $ units.

Such is the slope’s intuitive meaning. To discuss it algebraically, it will help to use “delta notation”, with which you might already be familiar: The expression $ x $ (read as “delta x”) refers to the change in x. For example, if x’s value has changed from -5 to 14, it has increased by 19 units; hence, $ x = 19 $ . (Similarly, if z’s value changes from 11 to 8, it has decreased by 3 units; hence, $ z = -3 $ .) And in general,

\[ \Delta x=(x^{\prime}s value after the change)-(x^{\prime}s value before the change). \]

Using this notation lets us express the formula for slope compactly. At right, we see a line with slope m. By our definition of slope above, when the line runs $ x $ units, it must rise by precisely $ m(x) $ units. Hence, between any two points on the line, we have $ y = m(x) $ . Dividing both sides by $ x $ yields $ m = y/x $ . That is,

\[ Slope=\frac{\Delta y}{\Delta x} \]

We can use this “rise over run” formula to compute a line’s slope from any two of its points.

Example. Find the slope of the line passing through $ (-1, 2) $ and $ (3, 7) $ .

Solution. The line’s slope is $ == $

Exercises

  1. Find the equation of…
  1. the vertical line through $ (-4,0) $ . b) the horizontal line through $ (0,2) $ . c) the x-axis. d) the y-axis. e) the vertical line through $ (a,0) $ . f) the horizontal line through $ (0,a) $ . g) the vertical line through $ (0,a) $ .
  1. Sketch the graphs of…
  1. x = 1 b) y = -2 c) y = 0 d) $ x = $

  1. Draw lines passing through the origin with the following slopes: 1, 2, 5, -1, -2, -5, 0, 1/2, -1/2. (Note that a line with negative slope falls instead of rises, while a line with zero slope is horizontal.)

  2. Suppose that a line has slope 3/4 and passes through (2, 5).

  1. The line will pass through a point whose x-coordinate is 9. What must the y-coordinate of that point be?

  2. The line will pass through a point whose x-coordinate is 11. What must the y-coordinate of that point be?

  1. If, over a three-hour period, the temperature T drops from $ 9^{} $ to $ -11^{} $ , what is $ T $ ?

  2. When a point moves from $ (2,3) $ to $ (7,12) $ , what is $ x $ ? What is $ y $ ? What is $ y/x $ ?

  3. When a point moves from (7, 12) to (2, 3), what is $ x $ ? What is $ y $ ? What is $ y/x $ ?

  4. Find the slope of the line passing through $ (2, 3) $ and $ (5, 15) $ .

  5. Find the slope of the line through $ (-200, 3010) $ and $ (-5, -6000) $ .

  6. If a line passes through points in the $ 2^{nd} $ and $ 4^{th} $ quadrants, what do we know about its slope?

Straight Lines: The Point-Slope Form

Boy, tell your story in a straight line, not in curves…

for to get a clear idea of the truth, one needs proofs and more proofs.

  • Cervantes, Don Quixote, Part II, Ch. XXVI

The solution to this next problem is the key to all that follows.

Problem 1. Derive the equation of the line through $ (2, 1) $ with slope -3.

Solution. We seek an equation that will be satisfied by the coordinates of all points on the line. To this end, we’ll let $ (x, y) $ be a variable point on the line. (Its variability is what lets it represent all the line’s points.)

The technique we’ll use to derive the line’s equation is to find two

distinct expressions for the line’s slope. (One of them will involve the variable point’s coordinates.) These expressions will represent the same thing, so they will, of course, be equal to one another. Thus, we’ll be able to put an equals sign between them. The result will be an equation satisfied by the variable point’s coordinates. That is, it will be satisfied by the coordinates of every point on the line. With this strategy in mind, the derivation is just four lines long:

On one hand, wherever the variable point $ (x,y) $ lies on the line, the slope from $ (2,1) $ to $ (x,y) $ is $ y/x = (-)/(-) $ . On the other, we’ve been given that this line’s slope is -3. Hence, wherever the variable point may be on the line, it always remains true that $ (y-1)/(x-2) = -3 $ . Therefore, this is the line’s equation.

We could mimic this argument whenever we want to find a line’s equation, but mathematicians are lazy. Instead of reinventing the wheel each time, let’s just solve this problem once and for all in the abstract.

Problem 2. Derive the equation of the line through $ (x_{0}, y_{0}) $ with slope m.

Solution. Let $ (x,y) $ be a variable point on the line. Wherever it lies, the slope from $ (x_{0},y_{0}) $ to $ (x,y) $ is $ y/x = ( - {0})/( - {0}) $ . But the line’s slope is also known to be m. Hence, wherever the variable point may roam on the line, it remains true that $ (y - y_{0})/(x - x_{0}) = m $ . Clearing fractions, we obtain the line’s equation in a fraction-free form:

\[ \boldsymbol{y}-\boldsymbol{y}_{0}=\boldsymbol{m}(x-x_{0}). \]

Lest you forget this tremendously important formula, I’ll summarize what we’ve learned here in a box.

The “Point-Slope” Formula for a Line

The equation of the line passing through point $ (x_{0}, y_{0}) $ with slope m is

\[ \boldsymbol{y}-\boldsymbol{y}_{0}=\boldsymbol{m}(\boldsymbol{x}-\boldsymbol{x}_{0}). \]

When using this point-slope formula to find a particular line’s equation, we typically apply a bit of algebraic polish at the end by distributing the slope and isolating y, as in the following example.

Problem 3. Find the equation of the line passing through $ (-3,4) $ with slope 2.

Solution. By the point-slope formula (with $ x_{0} = -3 $ , $ y_{0} = 4 $ , and m = 2), the line’s equation is

\[ y-4=2(x-(-3)). \]

After applying the customary algebraic polish, this becomes $ y = 2x + 10 $ .

You have just witnessed a remarkable mathematical pattern. Let us pause to acknowledge it and savor it. First, we solved a particular problem: We found the equation of the line through $ (2,1) $ with slope -3. This required the clever idea of equating two expressions for the slope, one of which involved a variable point’s coordinates. Recognizing this idea’s fruitfulness, we reconsidered the problem under general conditions, stripping away its particulars to expose its abstract essence: Find the equation of the line through point $ (x_{0},y_{0}) $ with slope m. Solving this abstract problem led us to discover the point-slope formula, which reduces the original particular problem (and any other one like it) to a mechanical triviality.

You’ve seen this pattern before. Solving our first substantial quadratic equation required ingenuity: the idea of “completing the square.” Recognizing that idea’s importance, we carried it up to the abstract air and applied it to the general quadratic, $ ax^{2} + bx + c = 0 $ , thereby deriving the quadratic formula.

We begin and end in particulars, but derive power from our flights into abstraction.

Exercise

  1. Find the equations of the lines described below. Also find the points where each line crosses the axes.

  1. Through $ (5,1) $ with slope 2.

  2. Through $ (4,-7) $ with slope -6.

  3. through $ (,-) $ with slope $ - $ .

  4. Through $ (-1, -2) $ and $ (3, 1) $ . [Hint: You haven’t been given the slope, but… you can find it.]

  5. Through $ (,) $ and $ (-,0) $ . f) Through $ (0,b) $ with slope m.

Straight Lines: The Slope-Intercept Form

This right line – The pathway for Christians to walk in! – say divines – The emblem of moral rectitude! – says Cicero – The best line! – say cabbage planters – is the shortest line, says Archimedes, which can be drawn from one given point to another.

  • Laurence Sterne, Tristram Shandy, Book VI, Chapter XL

The point-slope formula reveals that the line passing through $ (0, b) $ with a slope of m has a particularly simple equation: $ y = mx + b $ . We call this the “slope-intercept form” of the line, because its first parameter, m, represents the line’s slope, and its second, b, represents the line’s “y-intercept” (the place it crosses the y-axis).

The “Slope-Intercept” Form of a Line

The equation of the line with slope m and y-intercept b is

\[ \boldsymbol{y}=\boldsymbol{m}\boldsymbol{x}+\boldsymbol{b}. \]

We usually use the point-slope form to find a line’s equation, but we then go on to simplify the result and report it in slope-intercept form because of its clear geometric interpretation. When you see the equation y = 3x - 1, for example, you can picture the corresponding graph in your mind’s eye – a line through $ (0, -1) $ rising 3 units for every unit that it runs. Under certain circumstances, we can reverse the process and “read” a line’s equation directly off of its graph. For example,

Example 1. What is the equation of the line depicted at right?

Solution. Between the points at which the line crosses the axes, it “runs” 3 units and “rises” -2 units. Thus, its slope is -2/3. Its y-intercept is 2, so by the slope-intercept form, its equation must be

\[ \boldsymbol{y}=-\frac{2}{3}\boldsymbol{x}+\boldsymbol{2}. \]

Example 2. Fahrenheit depends linearly on Celsius. (That is, the graph of the formula that converts Celsius to Fahrenheit is a straight line.) Use this fact to derive the conversion formula.

Solution. Naturally, we’ll label our axes C and F instead of x and y, but this won’t change anything of importance. Two points determine a line, so if we can think of two points that lie on the line we seek, we can find its equation. Fortunately, there are two points that everyone knows: Water’s freezing point gives us one, (0, 32), and its boiling point gives us the other, (100, 212). All that remains is to find the equation of the line passing through them.

The figure shows that its slope is $ = = $ .

The line crosses the vertical axis at 32, so the line (and thus, the conversion formula) must be

\[ \boldsymbol{F}=\frac{9}{5}\boldsymbol{C}+3\boldsymbol{2}. \]

Exercises

  1. Find the equation of each of the following lines – and the points where each line crosses the axes.

  1. The line with slope 7 and y-intercept 2

  2. The line through $ (1,2) $ and $ (4,0) $

  3. The line through $ (2, -1/3) $ with slope 1

  4. The line with slope 7 and x-intercept 2

  1. In Example 2, you saw that the formula for converting Celsius to Fahrenheit is $ F = (9/5)C + 32 $ .

  1. For each degree that the Celsius scale increases, how many degrees does Fahrenheit increase?

  2. If the temperature rises by $ 5^{} $ C, what is the increase in degrees Fahrenheit?

  3. Find the formula for converting Fahrenheit to Celsius.

  4. Is there a temperature at which Fahrenheit and Celsius are equal? If so, what is it? If not, why not?

  1. Sketch graphs of the following equations and find where the graphs intersect the axes.
  1. $ y = x - 2 $ b) $ y = -2x + 5 $ c) $ 3x + 4y = 5 $ d) $ y = x $ e) x = -20
  1. Find the equations of the following lines:
a)

b)

c)

  1. Parallel lines have equal slopes. (This is fairly obvious, as you should convince yourself.) Using this fact…

  1. Find the equation of the line that passes through $ (3,2) $ and is parallel to the graph of $ y = -4x + 7 $ .

  2. Find the equation of the line through the origin that is parallel to the graph of $ 2x + 5y = 8 $ .

  1. Perpendicular lines have negative reciprocal slopes. For example, if a line’s slope is 2/5, then any line that is perpendicular to it has a slope of -5/2. Here’s a proof that this is true, followed by a couple of exercises.

Proof: If two lines are perpendicular, it’s clear that one line has a positive slope

(let’s call that slope p) and the other has a negative slope (which we’ll call n). *

We must prove that p = -1/n.

Draw a 1-unit horizontal line segment rightwards from the lines’ intersection. From its endpoint, draw a vertical line extending to the two original lines.

This creates a pair of similar right triangles.

The top one’s height is, by our definition of slope, p.

The bottom one’s height is $ -n. $ ^{†}

Similar triangles have the same proportions, so here, p/1 = 1/−n.

That is, p = -1/n, as claimed.

  1. Find the equation of the line passing through $ (3,2) $ and perpendicular to the graph of $ y = -4x + 7 $ .

  2. Find the equation of the line through the origin and perpendicular to the graph of $ 2x + 5y = 8 $ .

23. (Linear Equations in the form $ ax + by = c $ )

Sometimes it’s convenient to write linear equations in the form

\[ a x+b y=c. \]

One advantage of this form is that it lets us get rid of ugly fractional coefficients. For example, given the following linear equation in slope-intercept form,

\[ y=\frac{2}{3}x-\frac{1}{7}, \]

we can clear the fractions and rearrange the terms, putting it in this new form:

\[ 14x-21y=3. \]

Moreover, when a line’s equation is written in this form, we can read both of its intercepts off the equation. Consider 14x - 21y = 3. What is this line’s y-intercept? Well, where the line crosses the y-axis, x must be 0, and if we let x be 0 in this equation, then y = -1/7 (as you can surely calculate in your head). A similar thought process shows that the line’s x-intercept must be 3/14. As a result, it is supremely easy to graph a linear equation when it is written in this form: Just mark the two intercepts on the axes, and then draw the line that passes through them.

Use this method to graph the following lines:

\[ a)3x+6y=18\quad b)-3x-8y=48\quad c)5x+3y=10\quad d)2x+7y=3 \]

Pythagorean Gold Rush

Geometry has two great treasures: one is the theorem of Pythagoras, the other the division of a line into mean and extreme ratio. The first we may compare to a mass of gold, the second a precious jewel.

  • Johannes Kepler

Suppose the three squares in the figure at right are made of gold. You and a friend are going to share them. One of you will get the large square; the other, however, will get both the small and the medium square. Your magnanimous friend lets you decide which share you’ll get. Assuming you want to maximize your share of gold, which option will you choose for yourself? The large square? Or the other two squares?

Did you decide? Now read on.

Many people who are posed this problem are surprised to learn that it does not matter which option you choose, because the largest square contains exactly as much gold as the other two squares combined! This counterintuitive result follows directly from the most famous of all geometric theorems.

The Pythagorean Theorem

In any right triangle, the square on the hypotenuse is the sum of the squares on the legs.

Students too often meet the Pythagorean Theorem in an equivalent but lusterless algebraic form that obscures its connection to area. This is unfortunate, because thinking about the theorem in terms of area (as Pythagoras did) helps us recognize the Pythagorean Theorem as a genuinely surprising claim.

But why is this surprising theorem true?

There are many proofs of the Pythagorean Theorem. The following is one of the cleverest.

Proof of the Pythagorean Theorem

Given any right triangle, construct two identical squares each of whose edges are as long as the sum of the triangle’s legs. Paint the identical squares grey. Since the two squares are identical,

they have the same area.

Next, make eight identical copies of our given right triangle. Arrange these on the squares, as in the figures below, placing four triangles on each square, covering up some of the grey paint.

Clearly, the grey area left uncovered by triangles in the left figure equals the grey area left uncovered in the right figure. But the grey stuff on the left is the square on the right triangle’s hypotenuse. The grey stuff on the right is the sum of the squares on the right triangle’s legs! Since these are equal, and since this argument clearly works for any right triangle, we’ve proved the Pythagorean Theorem.

The Pythagorean Theorem relates the three sides of any right triangle; if we know any two sides, the Pythagorean Theorem will always give us the third.

Example. If a right triangle has a 12 cm hypotenuse, and a 10 cm leg, how long is its other leg?

Solution. Let x be the remaining leg’s length.

By the Pythagorean Theorem, we know that 122 = 102 + x2.

Solving this for x, we find that $ x = = 2 $ .

That is, the remaining leg of the right triangle is $ 2 $ cm long.

Exercises

  1. If a right triangle has a 15 inch hypotenuse, and a 1 inch leg, how long is its other leg?

  2. If the legs of a right triangle are 3 miles and 8 miles, how long is its hypotenuse?

  3. If the two longest sides of a triangle are 6 units and 4 units, does the Pythagorean Theorem tell us anything about the shortest side? If so, what does it tell us? If not, why not?

  4. If the two longest sides of a right triangle are 6 units and $ 2 $ units, does the Pythagorean Theorem tell us anything about the shortest side? If so, what does it tell us? If not, why not?

  5. Everyone knows that a rectangle with base b and height h has area bh.

Less well-known is the fact that a parallelogram with base b and height h also has area bh.*

Proof. In the figure at right, we cut a triangle from a parallelogram, then reattach it to the opposite side, turning the parallelogram into a rectangle. Since no area was created or destroyed during this operation, the original parallelogram’s area must equal that of the rectangle, which is, of course, bh.

Your problem: Prove that every triangle’s area is half the product of its base and height.

[Hint: To find the parallelogram’s area, we related it to another shape whose area we already knew (a rectangle). Use the same strategy with the triangle. Be sure your argument holds for all triangles, not just right triangles.]

  1. In this exercise, you’ll prove the Pythagorean Theorem in another way. The proof begins as follows:

Consider an arbitrary right triangle. Call its hypotenuse c and its legs a and b.

To prove the Pythagorean Theorem, we must prove that $ c^{2} = a^{2} + b^{2} $ .

To this end, arrange four copies of the right triangle as shown in the figure.

Your problem: Complete the proof.

[Hint: Compute the large square’s area in two different ways, then equate the expressions you obtain. After some algebraic manipulation, you’ll be able to rewrite your equation as $ c{2}=a{2}+b^{2} $ , thus proving the theorem.]

The Distance Formula

Viewed from a distance, everything is beautiful.

  • Tacitus

Distance is coordinate geometry’s lifeblood: A point’s coordinates specify its distances from the axes, and all the major geometric magnitudes (lengths, areas, volumes) are distances or products of distances. The following “distance formula” is therefore of paramount importance.

The Distance Formula.

The distance d between any two points may be computed (from their coordinates) as follows:

\[ d=\sqrt{(\Delta x)^{2}+(\Delta y)^{2}}. \]

Proof. We can view the line segment joining any two points in the plane as the hypotenuse of a right triangle, whose legs are parallel to the axes and have lengths $ |x| $ and $ |y| $ . The Pythagorean Theorem yields

\[ \begin{aligned}d^{2}&=|\Delta x|^{2}+|\Delta y|^{2}\\&=(\Delta x)^{2}+(\Delta y)^{2}.^{\dagger}\end{aligned} \]

Take the square root of both sides and we get the distance formula.

The distance formula is thus a disguised version of the Pythagorean Theorem – one of that theorem’s many masks. You’ll meet others in later courses.

Example. Find the distance between the points $ (-2, -11) $ and $ (10, -9) $ .

Solution. By the distance formula, the distance we seek is

\[ d=\sqrt{(\Delta x)^{2}+(\Delta y)^{2}}=\sqrt{(10-(-2))^{2}+(-9-(-11))^{2}}=\sqrt{148}=\mathbf{2}\sqrt{\mathbf{3}\mathbf{7}}. \]

Should you require a decimal approximation, a calculator will tell you that $ 2 $ .

You’ve already mastered the equation of a line. Now that you know the distance formula, you can also understand the equation of a circle, which we’ll derive after the following

Exercise

  1. Find the exact distances between the points in the following pairs:

  1. (2, 3) and (10, 9)

  2. $ (-1, 13) $ and $ (9, 10) $

\[ c)\left(3,-1/4\right)and\left(-2,13/2\right). \]

  1. $ (x_{1}, y_{1}) $ and $ (x_{2}, y_{2}) $

  2. (2, 5) and (2, 13) [You don’t need the distance formula here.]

  3. The proof above actually has a small flaw. (Did you notice it?) The problem is that certain line segments can’t be viewed as hypotenuses of right triangles whose legs are parallel to the axes. When does this happen? Does the distance formula still hold in those cases? If so, explain why. If not, provide a counterexample.

Circles

It is he that sitteth upon the circle of the earth,

and the inhabitants thereof are as grasshoppers.

  • Isaiah 40:22

Just as straightness is distinguished by a constant slope, so circularity is distinguished by a constant radius: All points on a circle are equidistant from its center. This defining property of a circle yields its equation.

Problem 1. Derive the equation of the unique circle with center $ (2, 1) $ and radius 3.

Solution. We must find an equation satisfied by the coordinates of all the circle’s points. To this end, let $ (x, y) $ be a variable point on the circle. (Its variability lets it represent all points on the circle.)

We’ll express the distance between $ (2,1) $ and $ (x,y) $ in two ways:

By the distance formula, it is √(x − 2)² + (y − 1)²

By the definition of a circle’s radius, it is 3.

Equating these two expressions, we get an equation satisfied by all

points $ (x,y) $ on the circle. We can polish that equation by squaring both sides, eliminating that unsightly square root, and yielding the circle’s equation in a clean form:

\[ (x-2)^{2}+(y-1)^{2}=9.\quad\blacklozenge \]

Now we’ll repeat this argument abstractly; we’ll derive the equation of the circle with center $ (x_{0}, y_{0}) $ and radius r. Once we have it, we’ll never need to derive the equation of a particular circle again: Instead, we’ll just plug the relevant information into a formula, which will produce the equation for us.

Problem 2. Derive the equation of the unique circle with center $ (x_{0}, y_{0}) $ and radius r.

Solution. We want an equation satisfied by the coordinates of all the circle’s points. To this end, let $ (x, y) $ be a variable point on the circle. (Its variability lets it to represent all points on the circle.)

We’ll express the distance between $ (x_{0}, y_{0}) $ and $ (x, y) $ two ways:

By the distance formula, it is √(x − x0)2 + (y − y0)2.

By the definition of a circle’s radius, it is r.

Equating these two expressions, we get an equation satisfied by all points $ (x, y) $ on the circle. By squaring both sides, we obtain the general equation of a circle, which we summarize in the box below.

The Equation of a Circle

The equation of the circle with center $ (x_{0}, y_{0}) $ and radius r is

\[ (\boldsymbol{x}-\boldsymbol{x_{0}})^{2}+(\boldsymbol{y}-\boldsymbol{y_{0}})^{2}=\boldsymbol{r}^{2} \]

Problem 3. Find the equation of the circle with center $ (-8,0) $ and radius 7.

Solution. By the boxed equation above (with $ x_{0} = -8 $ , $ y_{0} = 0 $ , r = 3), the circle’s equation is

\[ (x-(-8))^{2}+(y-0)^{2}=49. \]

After simplification, this becomes $ (x+8){2}+y{2}=49 $ .

The center and radius won’t always be handed to you on a silver platter. But if you are given sufficient information to find them, you can still produce the circle’s equation, as in this next example.

Problem 4. Find the equation of the circle that has a diameter joining the points $ (-1, 6) $ and $ (3, 4) $ .

Solution. To find this circle’s equation, we’ll need its center and radius. Its center, C, is the midpoint of the given diameter. The midpoint’s coordinates are the averages of the endpoint’s coordinates.

Thus, the circle’s center is $ (,)=(1,5) $ .

As for the radius, this is just the distance from the center to $ (3,4) $ . The distance formula tells us that this is

\[ \sqrt{(3-1)^{2}+(4-5)^{2}}=\sqrt{5}. \]

Now that we know the circle’s center and radius, we have its equation:

\[ (\boldsymbol{x}-\mathbf{1})^{2}+(\boldsymbol{y}-\mathbf{5})^{2}=\mathbf{5}.\quad\text{♦} \]

Exercises

  1. The so-called “unit circle” is the circle of radius 1 centered at the origin. What is its equation?

    1. Find the equation of the circle with center $ (-3, 2) $ and radius 4.

  1. Find the points where the circle in part (a) crosses the axes.

  2. Find the points where the circle in part (a) crosses the line x = -1.

  3. Find any ten points on the circle (including the ones you’ve found already).

    1. Find the equation of the circle with center $ (1, 1/2) $ and radius 1/2.

  1. Find the points where the circle in part (a) crosses the axes.

  2. Find the points where the circle in part (a) crosses the line y = 1/2.

[Think geometrically. You shouldn’t have to use an equation for this part.]

  1. Find the coordinates of any ten points on the circle (including the ones you’ve found already).
  1. Find the equation of each of the following circles. Also, graph the circles and label their key points (center and intersections with axes) with their coordinates.

  1. the circle that has a diameter joining the points $ (-2, 6) $ and $ (3, 4) $ .

  2. the circle that passes through $ (2, 5) $ and has center $ (3, -1) $ .

  3. the circle with diameter 7 centered at the origin.

  1. Graph the following equations, and give the coordinates of key points on each graph.
  1. $ x^{2} + y^{2} = 1 $ b) $ (x + 2)^{2} + (y - 3)^{2} = 25 $ c) $ x^{2} + (y + 1)^{2} = 5 $

  1. Find the equations of four different circles of radius $ $ that pass through (1, 1).

  2. A square’s vertices are $ (0, 0) $ , $ (0, 1) $ , $ (1, 0) $ , $ (1, 1) $ . A circle passes through all of them. Find its equation.

  3. Suppose that the part of the ground illuminated by a certain street lamp is a circular disc 80 feet in diameter. The lamp is located 30 feet east and 20 feet north of the intersection of those two well-known perpendicular streets, Horizontal Ave. and Vertical Lane.

  1. How many feet of Horizontal Ave. does the lamp illuminate?

  2. How many feet of Vertical Lane does the lamp illuminate?

  3. How many square feet of ground does the lamp illuminate?

  1. (Semicircles) Whenever we try to isolate a variable in a circle’s equation, we will end up with two equations. For example, if we massage

\[ (x-2)^{2}+(y+3)^{2}=5 \]

until y is isolated, we obtain

\[ y=-3\pm\sqrt{5-(x-2)^{2}}, \]

as you should verify. $ ^{*} $ These graphs of these two equations (one with the plus sign and one with the minus sign) are the top half and bottom half of the circle, respectively.

  1. Explain why the graph of the equation with the + is the circle’s top half.

  2. Now try isolating x. You will again obtain two equations. What are their graphs?

  3. Find the equation of the bottom half of the unit circle (i.e. the circle with center $ (0,0) $ and r = 1).

  4. Find the equation of the right half of the circle with center $ (-1, 2) $ and radius 6.

  5. What does the graph of $ y = - $ look like?

Six Basic Graphs

You should learn the shapes of the following graphs so well that they – like circles and lines – appear before your mind’s eye whenever you encounter their equations:

Thinking about each of these preceding equations for a few minutes should suffice to explain their graphs’ shapes. Do this before you move on. Then do the same for the next three graphs, whose shapes you should also learn:

Exercises

  1. Does the graph of $ y = $ ever cross the line y = 10? If so, at what point does it cross?

  2. Why does the graph of $ y = $ have points in the third quadrant, unlike the graph of $ y = $ ?

  3. To attain large y-values for either of the functions in the last problem, one needs to put in very large x-values.

In other words, $ y = $ and $ y = $ both grow very slowly.

Which of these two slowpokes grows faster than the other? Give some numerical evidence for your claim.

  1. Among $ y = x^{2} $ , $ y = x^{3} $ , and $ y = |x| $ , which grows fastest? Which grows slowest?

  2. Does the graph of y = 1/x ever touch either axis? How can you be certain?

  3. There is one point that lies on all six of the graphs above. What are its coordinates?

  4. Sketch the basic shapes of the graphs of $ y = x^{4} $ , $ y = x^{5} $ , $ y = x^{6} $ , and $ y = x^{7} $ .

  5. Describe the relationship between the graphs of y = x, y = -x, and $ y = |x| $ .

I ntersections

The intersections of two graphs are the points lying on both of them. A simple method to find them is based directly on the Fundamental Principle of Coordinate Geometry.

Example 1. Where does the line y = 2x intersect the ellipse $ x^{2} + 9y^{2} - 9 = 0 $ ?

Solution. Let $ (a, b) $ represent any point that lies on both graphs.

Since it lies on the $ ellipse’ $ ‘s graph, its coordinates must, by the FPCG, satisfy the $ ellipse’ $ ’s equation. Thus, we know that

\[ a^{2}+9b^{2}-9=0. \]

Similarly, since it lies on the line’s graph, the FPCG tells us that

\[ b=2a. \]

A-ha! We now have two equations in two unknowns, a situation we discussed in the last chapter. Recall the process: First, we’ll combine the two equations into a single equation in one unknown. From there, we can just follow our noses to the problem’s end. In this case, we can combine the equations very easily, by substituting the second equation into the first, which yields

\[ a^{2}+9(2a)^{2}-9=0. \]

Solving this, we get $ a = / $ . These are the intersections’ x-coordinates. Then, since these points lie on the line, we can find their y-coordinates by plugging each x-coordinate into the line’s equation, $ y = 2x $ . When we do that, we find that the y-coordinates are $ / $ .

Thus, the line intersects the ellipse at points (3/√37, 6/√37) and (−3/√37, −6/√37).

Such is the process for finding intersections. Its logic, I hope, is crystal clear. That said, we can speed the process up: From a strictly computational standpoint, introducing $ (a,b) $ into the problem is superfluous, and we can cut to the chase by combining the graphs’ equations without substituting a and b for x and y, proceeding to the solution from there. $ ^{†} $ Introducing a and b as an intersection point’s coordinates has psychological value, since it helps us understand the logic behind the technique, but in hindsight, once we’ve digested that logic, we see that it isn’t computationally necessary. We may therefore abandon those training wheels and proceed to the faster, simplified procedure.

We can summarize the process concisely if we use some new terminology: A system of two equations is a pair of equations for which we seek “simultaneous solutions” – i.e. points that satisfy both equations. To solve a system of two equations means to find all such points. $ ^{†} $

Finding the Intersections of Two Graphs.

To find the intersections of two graphs, we solve the system of their equations.

Solving a system of two equations in two unknowns is a three-step process. First, combine the equations via some sort of substitution so as to produce an equation in one unknown. Second, solve this equation. Third, substitute its solution back into one of the original equations and obtain the remaining unknown.

Example 2. Find the intersection points of the circle and line at right.

Solution. Because the circle is centered at the origin and its radius is 2, its equation is $ x^{2} + y^{2} = 4 $ . Since the line has slope 2 and y-intercept 0, its equation is y = 2x. We want to find the points whose coordinates satisfy both equations.

To find them, we’ll solve the system of these two equations.

First, we’ll combine the two equations into a single equation involving just one unknown. We can do this by substituting the second equation into the first, yielding $ x^{2} + (2x)^{2} = 4 $ .

Second, we solve this equation. Doing this (I’ll leave the details to you) yields $ x = / $ . These are the intersection points’ x-coordinates.

Finally, to obtain the corresponding y-coordinates, we substitute these x-coordinates back into the line’s equation, whereby we find that $ y = / $ .

We therefore conclude that the line and circle intersect at $ (, ) $ and $ (, ) $ .

Example 3. Find the intersection points of the two circles at right.

Solution. Clearly, the larger circle’s equation is $ x{2}+(y-2){2}=4 $ , while the smaller circle’s is $ x{2}+y{2}=1 $ .

First, we must combine these two equations in two unknowns into one equation in one unknown. We can accomplish this by rewriting the small circle’s equation as $ x^{2} = 1 - y^{2} $ , and substituting this into the large circle’s equation to get $ (1 - y^{2}) + (y - 2)^{2} = 4 $ .

Second, we solve this equation. Doing so (you should supply the details) reveals that y = 1/4. This tells us that the y-coordinate of both intersection points is 1/4.

Finally, we obtain the points’ x-coordinates by substituting y = 1/4 into the small circle’s equation, thus obtaining an equation in x alone: $ x^{2} + (1/4)^{2} = 1 $ . Solving it yields $ x = /4 $ .

Thus, the intersection points of the two circles are $ (/4, 1/4) $ and $ (-/4, 1/4) $ .

When solving intersection problems, you may spell out the details as I did in example 1, or make use of the accelerated procedure we used in examples 2 and 3. The choice is yours; either way, be sure you understand why that method works.

Exercises

  1. Sketch the graphs of the following equations, and find the point(s) where they intersect:

\[ a)y=3x-2,\quad y=-2x+3 \]

\[ \mathsf{b})5x+3y=3,\quad2x-7y=1 \]

\[ \mathsf{c})x^{2}+y^{2}=1,\ y=-\frac{2}{3}x \]

\[ \mathsf{d})\left(x-2\right)^{2}+\left(y+1\right)^{2}=3,\quad x+3y=3\qquad\mathsf{e})y=x^{3},\quad y=3x \]

\[ \mathsf{f})y=x^{2},\ y=2x+1 \]

\[ \mathfrak{g})x^{2}+y^{2}=1,\ y=x^{2} \]

  1. $ y=,y=x/2 $

\[ i)y=\sqrt{x},\quad x=4000000 \]

  1. $ y = $ , $ y = x + 1 $ [Hint: Review Exercise 39.]
  1. Mr. Anonymous tries to find the intersections of the lines $ y = 2x + 1 $ and $ y = 2x + 2 $ as follows:

First, he equates the two expressions for y, obtaining $ 2x + 1 = 2x + 2 $ .

He then subtracts 2x from both sides, obtaining the absurd statement 1 = 2.

What happened? What does this mean? Did Mr. Anonymous make a mistake somewhere? Discuss.

  1. Later, Mr. Anonymous attempts to find the intersections of $ x^{2} + y^{2} = 1 $ and $ y = x + 2 $ .

He begins by substituting the linear equation into the circle’s equation.

After some algebraic manipulation, he ends up with the quadratic equation $ 2x^{2} + 4x + 3 = 0 $ . He finds that this quadratic has no real solutions, as you should verify.

What does this lack of solutions mean geometrically?

  1. The figure at right shows the graphs of two lines with their corresponding equations.

  1. Do they cross? If so, where? If not, why not?

  2. Find the equation of the line passing through $ (0,0) $ and perpendicular to the upper line in the figure.

[Hint: Recall Exercise 22.]

  1. Where does the line whose equation you found in part (b) cross the lower line in the figure?
  1. The hyperbola in the figure at right is the graph of y = 1/x.

Find the coordinates of the four points at which it intersects the circle.

[Hint: You will end up with a $ 4^{th} $ -degree polynomial equation that you must solve. Don’t give up. Look closely and you will see that it is really a quadratic in disguise.]

  1. A line segment’s perpendicular bisector is the unique line that is perpendicular to the segment and passes through its midpoint. Around 300 B.C., Euclid proved that all three perpendicular bisectors of any triangle’s sides intersect at a single point. This point, moreover, turns out to be triangle’s circumcenter, the center of the unique circle (the triangle’s “circumcircle”) that passes through all three of the triangle’s vertices.

Your problem: Find the equation of the unique circle passing through the points…

  1. $ (0, 0) $ , $ (3, 0) $ , $ (0, 2) $

  2. $ (0, 0) $ , $ (4, 0) $ , $ (-2, 2) $

  3. $ (0, 0) $ , $ (2, 4) $ , $ (6, 2) $ .

I ntersections (encore)

One common technique for finding intersections is particularly well-suited to the equations of lines. The trick is simple and is built upon just two ideas.

We know, of course, that multiplying an equation’s sides by the same nonzero constant produces an equivalent equation – that is, an equation with the same solutions. But if two equations are satisfied by the same coordinate pairs, then they have the same graph. This observation yields our first big idea:

Multiplying the sides of an equation by the same nonzero constant preserves its graph.

Suppose, for example, that we’re working with the graph of $ 2x + 5y = 1 $ . If we realize that replacing this equation with $ 6x + 15y = 3 $ would be more convenient (for some algebraic reason or other), we can safely make this algebraic change without affecting the underlying geometry. $ ^{*} $

Next, consider a balanced scale, with objects of equal weight already resting on its two pans. If we add an additional pound of gold on one pan, and 16 ounces of gold on the other, the scale will obviously remain balanced, because adding equals to equals preserves the equality. Or, restated algebraically, if a = b and c = d, then $ a + c = b + d $ . (The same idea holds for subtraction.) This is our second big idea:

Adding (or subtracting) the sides of two valid equations produces a third valid equation.

Combining these two ideas leads to our new intersection-finding technique: Multiply each equation’s sides by well-chosen constants, then add or subtract the resulting equations. The key is to choose the constants so that the subsequent addition or subtraction eliminates one of the two variables. Let’s see an example.

Example. Find the point $ (x, y) $ where the lines $ 5x + 3y = 3 $ and 2x - 7y = 1 intersect.

Solution. First, multiply both sides of the first equations by 2 and both sides of the second by -5. The lines’ graphs are preserved, but their new equations, $ 10x + 6y = 6 $ and $ -10x + 35y = -5 $ , suit our purposes better, because adding their corresponding sides will eliminate x. Thanks to those equal but opposite $ 10x $ terms, adding yields an equation with just one variable: 41y = 1.

From this, we can read off the intersection’s y-coordinate: 1/41. Feeding this into either of the original equations yields its x-coordinate: 24/41. Thus, the lines intersect at $ (24/41, 1/41) $ .

Exercises

  1. In the preceding example, we “multiplied the equations” by 2 and -5, but other values would work, too. Think of another promising pair of numbers; use them to re-solve the problem, and verify that the solution is the same.

  2. Sometimes, it suffices to multiply just one equation’s sides before adding or subtracting. For example, try solving the system $ 2x + y = 3 $ and $ 5x + 3y = 3 $ by multiplying the first equation’s sides by 3, then subtracting.

  3. Use the new technique described above to find the intersections of the following pairs of lines.

  1. $ 3x + 5y = 2 $

$ 6x - 4y = 1 $

\[ \begin{aligned}&2x-7y=5\\&3x+7y=4\end{aligned} \]

\[ \begin{aligned}c)19x-3y&=4\quad&d)x+y&=1\\-4x-6y&=1\quad&2x-y&=1\end{aligned} \]