The Coordinate Plane

Section 6.1 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.

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Definition 6.1.1. 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 $y \gt 0 \text{.}$

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Quadrant II refers to points $(x,y)$ with $x \lt 0$ and $y \gt 0 \text{.}$

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Quadrant III refers to points $(x,y)$ with $x \lt 0$ and $y \lt 0 \text{.}$

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Quadrant IV refers to points $(x,y)$ with $x \gt 0$ and $y \lt 0 \text{.}$

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Points on an axis are not in any of the four quadrants.

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Example 6.1.2. 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.

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Solution.

Plotting:

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$\mathcal{P} $is in Quadrant II,
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$\mathcal{Q}$ is in Quadrant IV,
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$\mathcal{R}$ is in Quadrant III, and
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$\mathcal{S}$ is on the $y$-axis so is not in any of the four quadrants.
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Principle 6.1.3. Distance between points.

Suppose $\mathcal{P}(x_1,y_1)$ and $\mathcal{Q}(x_2,y_2)$ are two points in the coordinate plane.

The distance between $\mathcal{P}$ and $\mathcal{Q}$ is found using the distance formula

\begin{equation} \left(\text{distance between $\mathcal{P}$ and $\mathcal{Q}$}\right) = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}.\tag{6.1.1} \end{equation}

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Equation (6.1.1) 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: $(a-b)^2 = (b-a)^2\text{.}$

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Example 6.1.4. Using the Distance Formula.

Find the distance between $\mathcal{P}(-3,5)$ and $\mathcal{Q}(2,-4)$ .

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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*}

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Example 6.1.5. 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 (6.1.1),

\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{6.1.2} \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 $\sqrt{16}=4\text{.}$

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