Math1410 The Cartesian Coordinate Plane

Section 1.3 The Cartesian Coordinate Plane

As a warm-up for the discussions of vectors and three-dimensional geometry yet to come, we will make a quick review of the Cartesian Coordinate Plane. Imagine two real number lines crossing at a right angle at $0$ as shown in Figure 1.3.1.

The Cartesian coordinate plane. Horizontal axis labelled x; vertical axis labelled y. — Figure 1.3.1. The Cartesian coordinate plane

The horizontal number line is usually called the $\mathbf{x}$ axis while the vertical number line is usually called the $\mathbf{y}$ axis. As with the usual number line, we imagine these axes extending off indefinitely in both directions. Having two number lines allows us to locate the positions of points off of the number lines as well as points on the lines themselves.

For example, consider the point $P$ in Figure 1.3.2. To use the numbers on the axes to label this point, we imagine dropping a vertical line from the $x$-axis to $P$ and extending a horizontal line from the $y$-axis to $P\text{.}$ This process is sometimes called “projecting” the point $P$ to the $x$ (respectively $y$) axis. We then describe the point $P$ using the ordered pair $(2,-4)\text{.}$ The first number in the ordered pair is called the abscissa or $\mathbf{x}$ coordinate and the second is called the ordinate or $\mathbf{y}$ coordinate. Taken together, the ordered pair $(2,-4)$ comprise the Cartesian coordinates of the point $P\text{.}$ In practice, the distinction between a point and its coordinates is blurred; for example, we often speak of “the point $(2,-4)\text{.}$” We can think of $(2,-4)$ as instructions on how to reach $P$ from the origin $(0, 0)$ by moving $2$ units to the right and $4$ units downwards. Notice that the order in the ordered pair is important — if we wish to plot the point $(-4,2)\text{,}$ we would move to the left $4$ units from the origin and then move upwards $2$ units, as below on the right.

A point P with coordinates (2,-4) is plotted in the Cartesian plane, with rectangular guidelines. — Figure 1.3.2. Plotting points in Cartesian coordinates

Two points are plotted in the Cartesian plane, showing the difference between (2,-4) and (-4,2). — Figure 1.3.2. Plotting points in Cartesian coordinates

When we speak of the Cartesian Coordinate Plane, we mean the set of all possible ordered pairs $(x,y)$ as $x$ and $y$ take values from the real numbers. Below is a summary of important facts about Cartesian coordinates.

Key Idea 1.3.3. Important Facts about the Cartesian Coordinate Plane.

$(a,b)$ and $(c,d)$ represent the same point in the plane if and only if $a = c$ and $b = d\text{.}$
$(x,y)$ lies on the $x$-axis if and only if $y = 0\text{.}$
$(x,y)$ lies on the $y$-axis if and only if $x=0\text{.}$
The origin is the point $(0,0)\text{.}$ It is the only point common to both axes.

Example 1.3.4. Plotting points in the Cartesian Plane.

Plot the following points: $A(5,8)\text{,}$ $B\left(-\frac{5}{2}, 3\right)\text{,}$ $C(-5.8, -3)\text{,}$ $D(4.5, -1)\text{,}$ $E(5,0)\text{,}$ $F(0,5)\text{,}$ $G(-7,0)\text{,}$ $H(0, -9)\text{,}$ $O(0,0)\text{.}$

Solution.

To plot these points, we start at the origin and move to the right if the $x$ coordinate is positive; to the left if it is negative. Next, we move up if the $y$ coordinate is positive or down if it is negative. If the $x$-coordinate is $0\text{,}$ we start at the origin and move along the $y$-axis only. If the $y$-coordinate is $0$ we move along the $x$-axis only.

A plot showing all of the points for this example in the Cartesian plane. — Figure 1.3.5. Plotting points in Example 1.3.4

The axes divide the plane into four regions called quadrants. They are labelled with Roman numerals and proceed counterclockwise around the plane: see Figure 1.3.6.

Cartesian plane with four quadrants labelled — Figure 1.3.6. The four quadrants of the Cartesian plane

For example, $(1,2)$ lies in Quadrant I, $(-1,2)$ in Quadrant II, $(-1,-2)$ in Quadrant III and $(1,-2)$ in Quadrant IV. If a point other than the origin happens to lie on the axes, we typically refer to that point as lying on the positive or negative $x$ axis (if $y = 0$) or on the positive or negative $y$ axis (if $x = 0$). For example, $(0,4)$ lies on the positive $y$ axis whereas $(-117,0)$ lies on the negative $x$ axis. Such points do not belong to any of the four quadrants.

Distance in the Plane.

Another important concept in Geometry is the notion of length. If we are going to unite Algebra and Geometry using the Cartesian Plane, then we need to develop an algebraic understanding of what distance in the plane means. Suppose we have two points, $P\left(x_{0}, y_{0}\right)$ and $Q\left(x_{1}, y_{1}\right),$ in the plane. By the distance $d$ between $P$ and $Q\text{,}$ we mean the length of the line segment joining $P$ with $Q\text{.}$ (Remember, given any two distinct points in the plane, there is a unique line containing both points.) Our goal now is to create an algebraic formula to compute the distance between these two points. Consider the generic situation in Figure 1.3.7.(a).

Illustrating the distance between two points P and Q as a line segment — (a)

With a little more imagination, we can envision a right triangle whose hypotenuse has length $d$ as drawn in Figure 1.3.7.(b). From the latter figure, we see that the lengths of the legs of the triangle are $\left\lvert x_{1} - x_{0}\right\rvert$ and $\left\lvert y_{1} - y_{0}\right\rvert$ so the Pythagorean Theorem

en.wikipedia.org/wiki/Pythagorean_Theorem

gives us

\begin{align*} \left\lvert x_{1} - x_{0}\right\rvert^2 + \left\lvert y_{1} - y_{0}\right\rvert^2\amp = d^2\\ \left(x_{1} - x_{0}\right)^2 + \left(y_{1} - y_{0}\right)^2 \amp = d^2\text{.} \end{align*}

(Since the square of a number is always positive, we can drop the absolute value signs.) By extracting the square root of both sides of the second equation and using the fact that distance is never negative, we get the following result.

Key Idea 1.3.8. The Distance Formula.

The distance $d$ between the points $P\left(x_{0}, y_{0}\right)$ and $Q\left(x_{1}, y_{1}\right)$ is:

\begin{equation} d = \sqrt{ \left(x_{1} - x_{0}\right)^2 + \left(y_{1} - y_{0}\right)^2}\text{.}\tag{1.3.1} \end{equation}

It is not always the case that the points $P$ and $Q$ lend themselves to constructing such a triangle. If the points $P$ and $Q$ are arranged vertically or horizontally, or describe the exact same point, we cannot use the above geometric argument to derive the distance formula. It is left to the reader in Exercise Exercise 1.3.15 to verify Equation (1.3.1) for these cases.

Example 1.3.9. Distance between two points.

Find and simplify the distance between $P(-2,3)$ and $Q(1,-3)\text{.}$

Solution.

Using The Distance Formula, we have

\begin{align*} d \amp = \sqrt{\left(x_{1} - x_{0} \right)^2 + \left(y_{1} - y_{0} \right)^2}\\ \amp = \sqrt{ (1-(-2))^2 + (-3-3)^2}\\ \amp = \sqrt{9 + 36}\\ \amp = 3 \sqrt{5}\text{,} \end{align*}

so the distance is $3 \sqrt{5}\text{.}$

Example 1.3.10. Finding points at a given distance.

Find all of the points with $x$-coordinate $1$ which are $4$ units from the point $(3,2)\text{.}$

Solution.

We shall soon see that the points we wish to find are on the line $x=1\text{,}$ but for now we’ll just view them as points of the form $(1,y)\text{.}$

We require that the distance from $(3,2)$ to $(1,y)$ be $4\text{.}$ The Distance Formula, Equation (1.3.1), yields

\begin{align*} d \amp = \sqrt{\left(x_{1}-x_{0}\right)^2+\left(y_{1}-y_{0}\right)^2} \\ 4 \amp = \sqrt{(1-3)^2+(y-2)^2}\\ 4 \amp = \sqrt{4+(y-2)^2}\\ 4^2 \amp = \left(\sqrt{4+(y-2)^2}\right)^2 \quad \text{(squaring both sides)} \\ 16 \amp = 4+(y-2)^2\\ 12\amp = (y-2)^2\\ (y-2)^2 \amp = 12\\ y-2 \amp = \pm \sqrt{12} \quad \text{(extracting the square root)} \\ y \amp = 2 \pm 2\sqrt{3}\text{.} \end{align*}

We obtain two answers: $(1, 2 + 2 \sqrt{3})$ and $(1, 2-2 \sqrt{3}).$ The reader is encouraged to think about why there are two answers.

Showing a point (1,y) on the line x=1, a distance of 4 units from (3,2) — Figure 1.3.11. Diagram for Example 1.3.10

Related to finding the distance between two points is the problem of finding the midpoint of the line segment connecting two points. Given two points, $P\left(x_{0}, y_{0}\right)$ and $Q\left(x_{1}, y_{1}\right)\text{,}$ the midpoint $M$ of $P$ and $Q$ is defined to be the point on the line segment connecting $P$ and $Q$ whose distance from $P$ is equal to its distance from $Q\text{.}$

A line segment from P to Q, with the midpoint M marked. — Figure 1.3.12. The midpoint of a line segment

If we think of reaching $M$ by going “halfway over” and “halfway up” we get the following formula.

Key Idea 1.3.13. The Midpoint Formula.

The midpoint $M$ of the line segment connecting $P\left(x_{0}, y_{0}\right)$ and $Q\left(x_{1}, y_{1}\right)$ is

\begin{equation*} M = \left( \dfrac{x_{0} + x_{1}}{2} , \dfrac{y_{0} + y_{1}}{2} \right)\text{.} \end{equation*}

If we let $d$ denote the distance between $P$ and $Q\text{,}$ we leave it as Exercise 1.3.16 to show that the distance between $P$ and $M$ is $d/2\text{,}$ which is the same as the distance between $M$ and $Q\text{.}$ This suffices to show that Key Idea 1.3.13 gives the coordinates of the midpoint.

Example 1.3.14. Finding the midpoint of a line segment.

Find the midpoint of the line segment connecting $P(-2,3)$ and $Q(1,-3)\text{.}$

Solution.

Using The Midpoint Formula, we have

\begin{align*} M \amp = \left( \dfrac{x_{0}+x_{1}}{2}, \dfrac{y_{0}+y_{1}}{2} \right)\\ \amp = \left( \dfrac{(-2)+1}{2}, \dfrac{3+(-3)}{2} \right) = \left(- \dfrac{1}{2}, \dfrac{0}{2} \right)\\ \amp = \left(- \dfrac{1}{2}, 0 \right)\text{.} \end{align*}

The midpoint is $\left(- \frac{1}{2}, 0 \right)\text{.}$

We close with a more abstract application of the Midpoint Formula.

Example 1.3.15. An abstract midpoint problem.

If $a \neq b\text{,}$ prove that the line $y = x$ equally divides the line segment with endpoints $(a,b)$ and $(b,a)\text{.}$

Solution.

To prove the claim, we use The Midpoint Formula to find the midpoint.

\begin{align*} M \amp = \left( \dfrac{a+b}{2}, \dfrac{b+a}{2} \right)\\ \amp = \left( \dfrac{a+b}{2}, \dfrac{a+b}{2} \right)\text{.} \end{align*}

Since the $x$ and $y$ coordinates of this point are the same, we find that the midpoint lies on the line $y=x\text{,}$ as required.

Exercises Exercises

1.

Plot and label the points $A(-3, -7)\text{,}$ $B(1.3, -2)\text{,}$ $C(\pi, \sqrt{10})\text{,}$ $D(0, 8)\text{,}$ $E(-5.5, 0)\text{,}$ $F(-8, 4)\text{,}$ $G(9.2, -7.8)$ and $H(7, 5)$ in the Cartesian Coordinate Plane given below.

$The Cartesian coordinate plane. Horizontal axis labelled x; vertical axis labelled y.$

Exercise Group.

Find the distance $d$ between the points and the midpoint $M$ of the line segment which connects them.

2.

$(1,2)\text{,}$ $(-3,5)$

3.

$(3, -10)\text{,}$ $(-1, 2)$

4.

$\left( \dfrac{1}{2}, 4\right)\text{,}$ $\left(\dfrac{3}{2}, -1\right)$

5.

$\left(- \dfrac{2}{3}, \dfrac{3}{2} \right)\text{,}$ $\left(\dfrac{7}{3}, 2\right)$

6.

$\left( \dfrac{24}{5}, \dfrac{6}{5} \right)\text{,}$ $\left( -\dfrac{11}{5}, -\dfrac{19}{5} \right)\text{.}$

7.

$\left(\sqrt{2}, \sqrt{3}\right)\text{,}$ $\left(-\sqrt{8}, -\sqrt{12}\right)$

8.

$\left(2 \sqrt{45}, \sqrt{12} \right)\text{,}$ $\left(\sqrt{20}, \sqrt{27} \right)\text{.}$

9.

$(0, 0)\text{,}$ $(x, y)$

10.

Find all of the points of the form $(x, -1)$ which are $4$ units from the point $(3,2)\text{.}$

11.

Find all of the points on the $y$-axis which are $5$ units from the point $(-5,3)\text{.}$

12.

Find all of the points on the $x$-axis which are $2$ units from the point $(-1,1)\text{.}$

13.

Find all of the points of the form $(x,-x)$ which are $1$ unit from the origin.

14.

Let’s assume for a moment that we are standing at the origin and the positive $y$-axis points due North while the positive $x$-axis points due East. Our Sasquatch-o-meter tells us that Sasquatch is 3 miles West and 4 miles South of our current position. What are the coordinates of his position? How far away is he from us? If he runs 7 miles due East what would his new position be?

15.

Verify The Distance Formula for the cases when:

The points are arranged vertically. (Hint: Use $P(a, y_0)$ and $Q(a, y_1)\text{.}$)
The points are arranged horizontally. (Hint: Use $P(x_0, b)$ and $Q(x_1, b)\text{.}$)
The points are actually the same point. (You shouldn’t need a hint for this one.)

16.

Verify the Midpoint Formula by showing the distance between $P(x_1, y_1)$ and $M$ and the distance between $M$ and $Q(x_2, y_2)$ are both half of the distance between $P$ and $Q\text{.}$

17.

Show that the points $A\text{,}$ $\;B$ and $C$ below are the vertices of a right triangle.

(a)

$A(-3,2)\text{,}$ $\;B(-6,4)\text{,}$ and $C(1,8)$

(b)

$A(-3, 1)\text{,}$ $\;B(4, 0)$ and $C(0, -3)$

18.

Find a point $D(x, y)$ such that the points $A(-3, 1)\text{,}$ $\;B(4, 0)\text{,}$ $\;C(0, -3)$ and $D$ are the corners of a square. Justify your answer.

19.

The world is not flat. (There are those who disagree with this statement. Look them up on the Internet some time when you’re bored.) Thus the Cartesian Plane cannot possibly be the end of the story. Discuss with your classmates how you would extend Cartesian Coordinates to represent the three dimensional world. What would the Distance and Midpoint formulas look like, assuming those concepts make sense at all?

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