Cartesian Coordinates and Graphing

An ordered pair $(x,y)$ of real numbers can be represented by a point in the plane. Construct a pair of perpendicular lines, one horizontal (the x-axis), the other vertical (the y-axis). Locate the point on the plane corresponding to $(x,y)$ by starting at the origin --- the place where the axes cross --- and moving x units horizontally and y units vertically. If x is positive, you move to the right; if x is negative, you move to the left. Likewise, if y is positive, you move up; if y is negative, you move down.

x and y are the Cartesian coordinates of the point.


Example. Plot the points $(1,3)$ , $(3,1)$ , $(4,0)$ , $(0,-5)$ , and $(-2,-1)$ on a set of coordinate axes.

$$\hbox{\epsfysize=2in \epsffile{graph1.eps}}\quad\halmos$$


The midpoint of a segment is the point halfway between the two points. If the points are $(a,b)$ and $(c,d)$ , the midpoint is

$$\left(\dfrac{1}{2}(a + c), \dfrac{1}{2}(b + d)\right).$$

$$\hbox{\epsfysize=1.75in \epsffile{graph2.eps}}$$


Example. Find the midpoint of the segment joining the points $(2,5)$ and $(-3,7)$ .

$$\dfrac{1}{2}(2 + (-3)) = -\dfrac{1}{2} \quad\hbox{and}\quad \dfrac{1}{2}(5 + 7) = 6.$$

The midpoint is $\left(-\dfrac{1}{2},6\right)$ .


The distance between points $(a,b)$ and $(c,d)$ can be found using Pythagoras' theorem. It is

$$d = \sqrt{(a - c)^2 + (b - d)^2}.$$


Example. Find the distance between the points $(3,-7)$ and $(15,-12)$ .

$$d = \sqrt{(3 - 15)^2 + (-7 - (-12))^2} = \sqrt{(-12)^2 + 5^2} = \sqrt{144 + 25} = \sqrt{169} = 13.\quad\halmos$$


An equation involving x and y can be represented by a set of points in the plane --- the graph of the equation. The graph of an equation consists of all points whose coordinates $(x,y)$ satisfy the equation.


Example. Graph $y = x + 1$ .

Select some x-values and plug them into the equation to find the corresponding y-values. Then plot the resulting points.

$$\vbox{\offinterlineskip \halign{& \vrule # & \strut \hfil \quad # \quad \hfil \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr & x & & -3 & & -2 & & -1 & & 0 & & 1 & & 2 & & 3 & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr & y = x + 1 & & -2 & & -1 & & 0 & & 1 & & 2 & & 3 & & 4 & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} }} $$

$$\hbox{\epsfysize=2in \epsffile{graph3.eps}}\quad\halmos$$


Example. Graph $y = x^2$ .

$$\vbox{\offinterlineskip \halign{& \vrule # & \strut \hfil \quad # \quad \hfil \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr & x & & -2 & & -1 & & 0 & & 1 & & 2 & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr & $y = x^2$ & & 4 & & 1 & & 0 & & 1 & & 4 & \cr height2pt & \omit & & \omit & & \omit & & \omit & & \omit & & \omit & \cr \noalign{\hrule} }} $$

$$\hbox{\epsfysize=2in \epsffile{graph4.eps}}\quad\halmos$$


Example. Graph $x^2 + y^2 = 25$ .

$$\hbox{\epsfysize=2in \epsffile{graph5.eps}}\quad\halmos$$


The x-intercepts of an equation are the places where the graph crosses the x-axis. You can find the x-intercepts by setting $y = 0$ and solving for x.

The y-intercepts of an equation are the places where the graph crosses the y-axis. You can find the x-intercepts by setting $x = 0$ and solving for y.


Example. Find the x-intercepts and y-intercepts of $y = x^2 - 3x - 4$ .

To find the y-intercepts, set $x = 0$ :

$$y = 0^2 - 3\cdot 0 - 4 = -4.$$

To find the x-intercepts, set $y = 0$ :

$$0 = x^2 - 3x - 4, \quad 0 = (x - 4)(x + 1), \quad x = 4 \quad\hbox{or}\quad x = -1.\quad\halmos$$


Example. Find the x-intercepts and y-intercepts of $y = \dfrac{x + 3}{x -
   4}$ .

To find the y-intercepts, set $x = 0$ :

$$y = \dfrac{0 + 3}{0 - 4} = -\dfrac{3}{4}.$$

To find the x-intercepts, set $y = 0$ :

$$0 = \dfrac{x + 3}{x - 4}, \quad (x - 4)\cdot 0 = (x - 4)\cdot \dfrac{x + 3}{x - 4}, \quad 0 = x + 3, \quad x = -3.\quad\halmos$$


In what follows, I'll say two equations are the same if you can get from either one to the other using valid algebra.

You can often use symmetry as an aid in graphing equations.

The graph of an equation is symmetric about the y-axis if the equation is the same when x is replaced with $-x$ .


Example. $y =
   \dfrac{1}{x^2}$ is symmetric about the y-axis. If I replace x with $-x$ , I get

$$y = \dfrac{1}{(-x)^2} = \dfrac{1}{x^2},$$

which is the same as the original equation.

On the other hand, consider $y = x - 5$ . If I replace x with $-x$ , I get

$$y = (-x) - 5 = -x - 5.$$

This is not the same as the original equation, so the graph is not symmetric about the y-axis.


The graph of an equation is symmetric about the x-axis if the equation is the same when y is replaced with $-y$ .


Example. $x =
   y^2 + 5$ is symmetric about the x-axis. If I replace y with $-y$ , I get

$$x = (-y)^2 + 5 = y^2 + 5,$$

which is the same as the original equation.

On the other hand, consider $x + y = 3$ . If I replace y with $-y$ , I get

$$x + (-y) = 3 \quad\hbox{or}\quad x - y = 3.$$

This is not the same as the original equation, so the graph is not symmetric about the x-axis.


The graph of an equation is symmetric about the origin if the equation is the same when x is replaced with $-x$ and y is replaced with $-y$ .


Example. $xy =
   1$ is symmetric about the origin. If I replace x with $-x$ and y with $-y$ , I get

$$(-x)(-y) = 1 \quad\hbox{or}\quad xy = 1.$$

This is the same as the original equation.

On the other hand, consider $y = x^2$ . If I replace x with $-x$ and y with $-y$ , I get

$$-y = (-x)^2 \quad\hbox{or}\quad -y = x^2.$$

This is not the same as the original equation, so the graph is not symmetric about the origin.


An equation of the form

$$(x - a)^2 + (y - b)^2 = r^2$$

represents a circle of radius r whose center is the point $(a,b)$ .


Example. The equation

$$(x - 4)^2 + (y + 17)^2 = 25$$

represents a circle of radius $\sqrt{25}
   = 5$ with center $(4,-17)$ .


Example. Find an equation for the circle of radius 3 whose center is $(7,-10)$ .

$$(x - 7)^2 + (y + 10)^2 = 9.\quad\halmos$$


Example. Find the radius and the center of the circle $x^2 + y^2 - 6y = 55$ .

I need to add a number to $y^2 - 6y$ to make a perfect square.

$$\dfrac{-6}{2} = -3 \quad\hbox{and}\quad (-3)^2 = 9,$$

so I add 9 to both sides:

$$x^2 + y^2 - 6y = 55, \quad x^2 + y^2 - 6y + 9 = 55 + 9, \quad x^2 + (y - 3)^2 = 64.$$

Since $\sqrt{64} = 8$ , the radius is 8. The center is $(0,3)$ .


Example. Find the radius and the center of the circle $x^2 + 2x + y^2 - 10y = 10$ .

I need to add numbers to $x^2 + 2x$ and to $y^2 - 10y$ to make perfect squares.

$$\dfrac{2}{2} = 1 \quad\hbox{and}\quad 1^2 = 1.$$

$$\dfrac{-10}{2} = -5 \quad\hbox{and}\quad (-5)^2 = 25.$$

Thus, I need to add 1 to the first expression and 25 to the second. Since I'm adding $1 + 25 = 26$ to the left side, I must add 26 to the right side of the equation as well.

$$x^2 + 2x + y^2 - 10y = 10, \quad x^2 + 2x + 1 + y^2 - 10y + 25 = 10 + 26, \quad (x + 1)^2 + (y - 5)^2 = 36.$$

The center is $(-1,5)$ and the radius is 6.


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