Coordinates, graphs, and circles
Graphs turn equations into pictures. This page sets up the coordinate plane, gives formulas for the distance and the midpoint between two points, and shows how to graph an equation. It ends with the equation of a circle.
The coordinate plane
Draw two number lines that cross at right angles at their zero points. The horizontal one is the ${x}$-axis, and the vertical one is the ${y}$-axis. They cross at the origin ${O}$. Together they make a coordinate plane (also called a rectangular, or Cartesian, coordinate system).
Each point ${P}$ in the plane matches one ordered pair ${(a,b)}$. The first number, ${a}$, is the ${x}$-coordinate: how far ${P}$ is to the right of the ${y}$-axis (negative means left). The second number, ${b}$, is the ${y}$-coordinate: how far ${P}$ is above the ${x}$-axis (negative means below). The order matters: ${(3,2)}$ and ${(2,3)}$ are different points.
The axes split the plane into four quadrants, numbered I to IV counterclockwise:
- Quadrant I: ${x>0}$ and ${y>0}$, like ${A(3,2)}$.
- Quadrant II: ${x<0}$ and ${y>0}$, like ${B(-2,4)}$.
- Quadrant III: ${x<0}$ and ${y<0}$, like ${C(-3,-1)}$.
- Quadrant IV: ${x>0}$ and ${y<0}$, like ${D(2,-3)}$.
A point on an axis, like ${(0,5)}$, is not in any quadrant.
The distance formula
To find the distance between two points, draw a right triangle, as in the figure for ${P(-2,3)}$ and ${Q(4,-5)}$. The corner of the triangle is ${(-2,-5)}$.
- The horizontal leg runs from ${(-2,-5)}$ to ${Q}$. Its length is ${4-(-2)=6}$.
- The vertical leg runs from ${P}$ down to ${(-2,-5)}$. Its length is ${3-(-5)=8}$.
By the Pythagorean theorem, ${d^2=6^2+8^2=100}$, so ${d=10}$. In general, the distance between ${P_1(x_1,y_1)}$ and ${P_2(x_2,y_2)}$ is
${d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}}$
The order of the points does not matter, because each difference is squared.
The midpoint formula
The midpoint of a segment is the point halfway between its ends. Its coordinates are the averages of the endpoints' coordinates:
$M=\left(\dfrac{x_1+x_2}{2},\ \dfrac{y_1+y_2}{2}\right)$
Example 1: Let ${A(1,-2)}$ and ${B(7,6)}$.
(a) Find the distance between ${A}$ and ${B}$.
(b) Find the midpoint of segment ${AB}$.
Solution:
(a) Put the coordinates into the distance formula:
$\begin{align*}&d\\&=\sqrt{(7-1)^2+(6-(-2))^2}\\&=\sqrt{6^2+8^2}\\&=\sqrt{100}=10\end{align*}$
(b) Average the ${x}$-coordinates, and average the ${y}$-coordinates:
$\begin{align*}&M\\&=\left(\dfrac{1+7}{2},\ \dfrac{-2+6}{2}\right)\\&=(4,2)\end{align*}$
Check: ${M}$ should be ${5}$ units from each end. From ${A}$: ${\sqrt{3^2+4^2}=5}$. It checks.
Graphs of equations
A solution of an equation in ${x}$ and ${y}$ is an ordered pair that makes it true. For example, ${(3,5)}$ is a solution of ${y=x^2-4}$, because ${5=3^2-4}$. The graph of the equation is the set of all its solutions, drawn as points in the plane.
To sketch a graph by hand, make a table of solutions, plot them, and join them with a smooth curve.
Two kinds of points help the most. An ${x}$-intercept is where the graph crosses the ${x}$-axis, so ${y=0}$ there. A ${y}$-intercept is where it crosses the ${y}$-axis, so ${x=0}$ there.
- To find the ${x}$-intercepts, set ${y=0}$ and solve for ${x}$.
- To find the ${y}$-intercepts, set ${x=0}$ and solve for ${y}$.
Example 2: Find the intercepts of ${y=x^2-4}$, and sketch its graph.
Solution:
${x}$-intercepts: Set ${y=0}$:
$\begin{align*}x^2-4&=0\\(x+2)(x-2)&=0\end{align*}$
So the ${x}$-intercepts are ${-2}$ and ${2}$.
${y}$-intercept: Set ${x=0}$: ${y=0^2-4=-4}$.
More points: When ${x=\pm 1}$, ${y=-3}$. When ${x=\pm 3}$, ${y=5}$. Plot these and join them:
Symmetry
The graph in Example 2 is a mirror image of itself across the ${y}$-axis. This is called symmetry, and it halves the work of graphing. There are three tests:
| The graph is symmetric about | if this gives the same equation |
|---|---|
| the ${y}$-axis | replace ${x}$ by ${-x}$ |
| the ${x}$-axis | replace ${y}$ by ${-y}$ |
| the origin | replace both ${x}$ by ${-x}$ and ${y}$ by ${-y}$ |
Example 3: Test each graph for symmetry.
(a) ${y=x^2-4}$
(b) ${y=x^3-x}$
Solution:
(a) Replace ${x}$ by ${-x}$: ${y=(-x)^2-4=x^2-4}$. This is the same equation, so the graph is symmetric about the ${y}$-axis.
(b) Replace ${x}$ by ${-x}$ and ${y}$ by ${-y}$:
$\begin{align*}-y&=(-x)^3-(-x)\\-y&=-x^3+x\\y&=x^3-x\end{align*}$
This is the same equation, so the graph is symmetric about the origin. (Turn it upside down, and it looks the same.)
Circles
A circle is the set of all points at a fixed distance ${r}$, the radius, from a fixed point ${(h,k)}$, the center. A point ${(x,y)}$ is on the circle when its distance from ${(h,k)}$ is ${r}$. By the distance formula, ${\sqrt{(x-h)^2+(y-k)^2}=r}$. Squaring both sides gives the standard equation of a circle:
${(x-h)^2+(y-k)^2=r^2}$
The circle with center ${(0,0)}$ and radius ${1}$, with equation ${x^2+y^2=1}$, is called the unit circle.
Example 4: Find the equation of the circle with center ${(-1,4)}$ and radius ${3}$.
Solution:
Here ${h=-1}$, ${k=4}$, and ${r=3}$. Be careful with the sign: ${x-h=x-(-1)=x+1}$.
${(x+1)^2+(y-4)^2=9}$
Example 5: Find the center and radius of the circle
${x^2+y^2-6x+4y-3=0}$.
Solution:
Complete the square twice: once for the ${x}$ terms and once for the ${y}$ terms. First group them, and move the constant:
${(x^2-6x)+(y^2+4y)=3}$
Half of ${-6}$ is ${-3}$, and ${(-3)^2=9}$. Half of ${4}$ is ${2}$, and ${2^2=4}$. Add ${9}$ and ${4}$ to both sides:
$\begin{align*}(x^2-6x+9)+(y^2+4y+4)&=3+9+4\\(x-3)^2+(y+2)^2&=16\end{align*}$
So the center is ${(3,-2)}$, and the radius is ${\sqrt{16}=4}$.
Summary
- A point is an ordered pair ${(x,y)}$. The axes split the plane into four quadrants.
- Distance: ${d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}}$. Midpoint: average the coordinates.
- To find ${x}$-intercepts, set ${y=0}$. To find ${y}$-intercepts, set ${x=0}$.
- Symmetry tests: replace ${x}$ by ${-x}$ (${y}$-axis), ${y}$ by ${-y}$ (${x}$-axis), or both (origin).
- A circle with center ${(h,k)}$ and radius ${r}$ has the equation ${(x-h)^2+(y-k)^2=r^2}$. To find the center and radius, complete the square in ${x}$ and in ${y}$.