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Precalculus JumpStart

Section 5.1 The Coordinate Plane

The Cartesian coordinate plane is formed by drawing perpendicular coordinate axes starting at a point called the origin. The location of a point \(\mathcal{P}\) in the plane is determined by an ordered pair of real numbers \((x,y)\) called its coordinates. The first coordinate \(x\) of the pair determines the location of \(\mathcal{P}\) along the horizontal axis and the second coordinate \(y\) of the pair determines its location along the vertical axis.
The origin \(\mathcal{O}\) has coordinates \((0,0)\text{.}\) Points on the \(x\)-axis have coordinates \(\mathcal{Q}(x,0)\text{,}\) with the second coordinate zero. Similarly, points on the \(y\)-axis have coordinates \(\mathcal{R}(0,y)\text{,}\) with the first coordinate zero.

Definition 5.1. The Four Quadrants.

The coordinate axes divides the plane into four quadrants labeled by roman numerals counter-clockwise:
  • Quadrant I refers to points \((x,y)\) with \(x \gt 0\) and \(y \gt 0 \text{.}\)
  • Quadrant II refers to points \((x,y)\) with \(x \lt 0\) and \(y \gt 0 \text{.}\)
  • Quadrant III refers to points \((x,y)\) with \(x \lt 0\) and \(y \lt 0 \text{.}\)
  • Quadrant IV refers to points \((x,y)\) with \(x \gt 0\) and \(y \lt 0 \text{.}\)
Points on an axis are not in any of the four quadrants.

Example 5.2. Plotting Points.

Plot the points \(P(-3,5)\text{,}\) \(\mathcal{Q}(2,-4)\text{,}\) \(\mathcal{R}(-1.5,-2.5)\text{,}\) and \(\mathcal{S}(0,4)\) on the coordinate plane and determine the quadrant of each point, if applicable.
Solution.
Plotting:
  • \(\mathcal{P} \)is in Quadrant II,
  • \(\mathcal{Q}\) is in Quadrant IV,
  • \(\mathcal{R}\) is in Quadrant III, and
  • \(\mathcal{S}\) is on the \(y\)-axis so is not in any of the four quadrants.
Equation (5.1) is a “Pythagorean” expression: sum the squares of the differences of corresponding coordinates and then take the positive square root of this. Note that the order of subtraction is irrelevant due to the squaring: \((a-b)^2 = (b-a)^2\text{.}\)

Example 5.4. Using the Distance Formula.

Find the distance between \(\mathcal{P}(-3,5)\) and \(\mathcal{Q}(2,-4)\) .
Solution.
\begin{align*} \left(\text{distance}\right) \amp=\quad \sqrt{(2-(-3))^2 + (-4-5)^2} \\ \amp=\quad \sqrt{5^2 + 9^2} \\ \amp=\quad \sqrt{25 + 81} \\ \amp=\quad \sqrt{106} \\ \amp\approx\quad 10.3 \end{align*}

Example 5.5. The Equation of a Circle.

Suppose \(\mathcal{C}(a,b)\) is a point in the plane and \(r\gt 0\) is a positive real number. The circle centered at \(\mathcal{C}\) with radius \(r\) consists of all points \(\mathcal{P}(x,y)\) satisfying
\begin{equation*} \left(\text{distance between $\mathcal{P}$ and $\mathcal{Q}$}\right) = r \end{equation*}
From the Distance Formula (5.1),
\begin{equation*} \sqrt{(x-a)^2+(y-b)^2}= r. \end{equation*}
It is convenient to square both sides of this equation to remove the radical and obtain the following equation of a circle
\begin{equation} (x-a)^2+(y-b)^2=r^2.\tag{5.2} \end{equation}
For example
\begin{equation*} (x-2)^2 + (y+3)^2 = 16 \end{equation*}
is the equation of the circle with center \(\mathcal{C}(2,-3)\) and radius \(\sqrt{16}=4\text{.}\)