The Cartesian coordinate system identifies the location of every point in a plane with an ordered pair.
Example4.11.1
On paper, sketch a Cartesian coordinate system with units, and then plot the following points: \((3,2),(-5,-1),(0,-3),(4,0)\text{.}\)
Subsection4.11.2Review of Graphing Equations
Graphing Equations
To graph an equation with two variables \(x\) and \(y\text{,}\) we can choose some reasonable \(x\)-values, then calculate the corresponding \(y\)-values, and then plot the \((x,y)\)-pairs as points. For many (not-so-complicated) algebraic equations, connecting those points with a smooth curve will produce an excellent graph.
Example4.11.2
Graph the equation \(y=-2x+5\text{.}\)
\(x\)
\(y=-2x+5\)
Point
\(-2\)
\(\phantom{-2(-2)+5=\substitute{9}}\)
\(\phantom{(-2,9)}\)
\(-1\)
\(\phantom{-2(-1)+5=\substitute{7}}\)
\(\phantom{(-1,7)}\)
\(0\)
\(\phantom{-2(0)+5=\substitute{5}}\)
\(\phantom{(0,5)}\)
\(1\)
\(\phantom{-2(1)+5=\substitute{3}}\)
\(\phantom{(1,3)}\)
\(2\)
\(\phantom{-2(2)+5=\substitute{1}}\)
\(\phantom{(2,1)}\)
\(x\)
\(y=-2x+5\)
Point
\(-2\)
\(-2(-2)+5=\substitute{9}\)
\((-2,9)\)
\(-1\)
\(-2(-1)+5=\substitute{7}\)
\((-1,7)\)
\(0\)
\(-2(0)+5=\substitute{5}\)
\((0,5)\)
\(1\)
\(-2(1)+5=\substitute{3}\)
\((1,3)\)
\(2\)
\(-2(2)+5=\substitute{1}\)
\((2,1)\)
(a)Set up the table(b)Complete the table
Figure4.11.3Making a table for \(y=-2x+5\)
We use points from the table to graph the equation. First, plot each point carefully. Then, connect the points with a smooth curve. Here, the curve is a straight line. Lastly, we can communicate that the graph extends further by sketching arrows on both ends of the line.
(a)Use points from the table(b)Connect the points in whatever pattern is apparent
Figure4.11.4Graphing the Equation \(y=-2x+5\)
Subsection4.11.3Review of Two-Variable Data and Rate of Change
Exploring Two-Variable Data and Rate of Change
For a linear relationship, by its data in a table, we can see the rate of change (slope) and the line's \(y\)-intercept, thus writing the equation.
Example4.11.5
Write an equation in the form \(y=\ldots\) suggested by the pattern in the table.
\(x\)
\(y\)
\(0\)
\(-4\)
\(1\)
\(-6\)
\(2\)
\(-8\)
\(3\)
\(-10\)
We consider how the values change from one row to the next. From row to row, the \(x\)-value increases by \(1\text{.}\) Also, the \(y\)-value decreases by \(2\) from row to row.
\(x\)
\(y\)
\(0\)
\(-4\)
\({}+1\rightarrow\)
\(1\)
\(-6\)
\(\leftarrow{}-2\)
\({}+1\rightarrow\)
\(2\)
\(-8\)
\(\leftarrow{}-2\)
\({}+1\rightarrow\)
\(3\)
\(-10\)
\(\leftarrow{}-2\)
Since row-to-row change is always \(1\) for \(x\) and is always \(-2\) for \(y\text{,}\) the rate of change from one row to another row is always the same: \(-2\) units of \(y\) for every \(1\) unit of \(x\text{.}\)
We know that the output for \(x = 0\) is \(y = -4\text{.}\) And our observation about the constant rate of change tells us that if we increase the input by \(x\) units from \(0\text{,}\) the ouput should decrease by \(\overbrace{(-2)+(-2)+\cdots+(-2)}^{x\text{ times}}\text{,}\) which is \(-2x\text{.}\) So the output would be \(-4-2x\text{.}\)
So the equation is \(y=-2x-4\text{.}\)
Subsection4.11.4Review of Slope
Slope
When \(x\) and \(y\) are two variables where the rate of change between any two points is always the same, we call this common rate of change the slope. Since having a constant rate of change means the graph will be a straight line, it's also called the slope of the line.
We can find a line's slope by drawing a slope-triangle on the line's graph, and then using the formula
\begin{equation}
m=\frac{\text{change in $y$}}{\text{change in $x$}}=\frac{\Delta y}{\Delta x}\tag{4.11.1}
\end{equation}
Example4.11.6
Find the slope of the line in the following graph.
We picked two points on the line, and then drew a slope triangle. Next, we will do:
A line's equation in slope-intercept form looks like \(y=mx+b\text{,}\) where \(m\) is the line's slope, and \(b\) is the line's \(y\)-intercept.
We can use a line's \(y\)-intercept and slope triangles to graph it.
Example4.11.7
Graph the line \(y=-\frac{2}{3}x+10\text{.}\)
(a)First, plot the line's \(y\)-intercept, \((0,10)\text{.}\)(b)The slope is \(-\frac{2}{3}=\frac{-2}{3}=\frac{2}{-3}\text{.}\) So we can try using a “run” of \(3\) and a “rise” of \(-2\) or a “run” of \(-3\) and a “rise” of \(2\text{.}\)(c)Arrowheads and labels are encouraged.
Figure4.11.8Graphing \(y=-\frac{2}{3}x+10\)
Writing a Line's Equation in Slope-Intercept Form Based on Graph
Given a line's graph, we can identify its \(y\)-intercept, and then find its slope by a slope triangle. With a line's slope and \(y\)-intercept, we can write its equation in the form of \(y=mx+b\text{.}\)
Example4.11.9
Find the equation of the line in the graph.
(a)Graph of a line(b)Identify the line's \(y\)-intercept, \(10\text{.}\)(c)Identify the line's slope by a slope triangle. Note that we can pick any two points on the line to create a slope triangle. We would get the same slope: \(-\frac{2}{3}\)
Figure4.11.10Find the equation of the line in the graph.
With the line's slope \(-\frac{2}{3}\) and \(y\)-intercept \(10\text{,}\) we can write the line's equation in slope-intercept form: \(y=-\frac{2}{3}x+10\text{.}\)
Subsection4.11.6Review of Point-Slope Form
Point-Slope Form
A line's point-slope form equation is in the form of \(y=m\left(x-x_0\right)+y_0\text{,}\) where \(m\) is the line's slope, and \((x_0,y_0)\) is a point on the line.
Example4.11.11
A line passes through \((-6,0)\) and \((9,-10)\text{.}\) Find this line's equation in both point-slope and slope-intercept form.
Solution: We will use the slope formula (4.4.3) to find the slope first. After labeling those two points as \((\overset{x_1}{-6},\overset{y_1}{0}) \text{ and } (\overset{x_2}{9},\overset{y_2}{-10})\text{,}\) we have:
Now the point-slope equation looks like \(y=-\frac{2}{3}(x-x_0)+y_0\text{.}\) Next, we will use \((9,-10)\) and substitute \(x_0\) with \(9\) and \(y_0\) with \(-10\text{,}\) and we have:
A line's equation in standard form looks like \(Ax+By=C\text{.}\) We need to convert a line's equation from standard form to slope-intercept form, and vice versa.
Examples
Converting from Standard Form to Slope-Intercept Form
The line's equation in standard form is \(2x+3y=6\text{.}\)
To graph a line in standard form, we could first change it to slope-intercept form, and then graph the line by its \(y\)-intercept and slope triangles. A second method is to graph the line by its \(x\)-intercept and \(y\)-intercept.
Example4.11.12
Graph \(2x-3y=-6\) using its intercepts. And then use the intercepts to calculate the line's slope.
So the line's \(y\)-intercept is \((0,2)\text{.}\)
With both intercepts' coordinates, we can graph the line:
Figure4.11.13Graph of \(2x-3y=-6\)
Subsection4.11.8Review of Horizontal, Vertical, Parallel, and Perpendicular Lines
Horizontal, Vertical, Parallel, and Perpendicular Lines
A horizontal line's equation looks like \(y=k\text{,}\) while a vertical line's equation looks like \(x=h\text{.}\) The following figure has graphs of a horizontal line and a vertical line.
Figure4.11.14Graphs of \(y=3\) and \(x=5\)
Parallel and Perpendicular Lines
Examples
Two lines are parallel if and only if they have the same slope.
Line \(m\)'s equation is \(y=-2x+20\text{.}\) Line \(n\) is parallel to \(m\text{,}\) and line \(n\) also passes the point \((4,-3)\text{.}\) Find line \(n\)'s equaiton.
Since parallel lines have the same slope, line \(n\)'s slope is also \(-2\text{.}\) Since line \(n\) also passes the point \((4,-3)\text{,}\) we can write line \(n\)'s equation in point-slope form:
Two lines are perpendicular if and only if the product of their slopes is \(-1\text{.}\)
Line \(m\)'s equation is \(y=-2x+20\text{.}\) Line \(n\) is perpendicular to \(m\text{,}\) and line \(n\) also passes the point \((4,-3)\text{.}\) Find line \(n\)'s equaiton.
Since line \(m\) and \(n\) are perpendicular, the product of their slopes is \(-1\text{.}\) Because line \(m\)'s slope is given as \(-2\text{,}\) we can find line \(n\)'s slope is \(\frac{1}{2}\text{.}\)
Since line \(n\) also passes the point \((4,-3)\text{,}\) we can write line \(n\)'s equation in point-slope form:
Subsection4.11.9Review of Linear Inequalities in Two Variables
Linear Inequalities in Two Variables
When we graph lines like \(y=2x+1\text{,}\) we are graphing points which satisfy the relationship \(y=2x+1\text{,}\) like \((0,1), (1,3), (2,5)\text{,}\) etc. Similarly, we can graph linear inequalities like \(y\gt2x+1\) by plotting all points which satisfies the inequality, like \((0,2), (0,3), (1,4), (1,5)\text{,}\) etc. All these points form a region, instead of a line.
Example4.11.15
Graph \(y\gt2x+1\text{.}\)
There are two steps to graph an inequality.
Graph the line \(y=2x+1\text{.}\) Because the inequality symbol is \(\gt\text{,}\) (instead of \(\ge\)) the line should be dashed (instead of solid).
Next, we need to decide whether to shade the region above \(y=2x+1\) or below it. We will choose a point to test whether \(y\gt2x+1\) is true. As long as the line doesn't cross \((0,0)\text{,}\) we will use \((0,0)\) to test, because the number \(0\) is the easiest number for calculation.
\begin{gather*}
y\gt2x+1\\
0\stackrel{?}{\gt}2(0)+1\\
0\stackrel{?}{\gt}1
\end{gather*}
Because \(0\gt1\) is not true, the point \((0,0)\) is not a solution and should not be shaded. As a result, we shade the region without \((0,0)\text{.}\)
Figure4.11.16Step 1 of graphing \(y\gt2x+1\)Figure4.11.17Step 2 of graphing \(y\gt2x+1\)
Subsection4.11.10Exercises
1
Sketch the points \((8,2)\text{,}\) \((5,5)\text{,}\) \((-3,0)\text{,}\) \(\left(0,-\frac{14}{3}\right)\text{,}\) \((3,-2.5)\text{,}\) and \((-5,7)\) on a Cartesian plane.
