Introduction to the Coordinate System and T-Chart
In algebra and later classes, you will work with a Coordinate System where you’ll graph relations and functions. This is also called a Cartesian Plane.
To graph, put the $ x$’s (the independent variable) on the bottom and the $ y$’s (the dependent variables) up on the side like this. Don’t worry about why you are graphing this way; someone just decided to do it this way. First, go back and forth with the $ x$’s, and then up and down with the $ y$’s; the point is written like $ \left( {x,y} \right)$. (Remember that you learned to crawl first – back and forth – and then you learned to walk – which is more like going up and down.)
The $ \left( {x,y} \right)$ point is also called an ordered pair. For the point $ \left( {4,3} \right)$ below, we went over 4 with the $ x$’s and up 3 with the $ y$’s. The $ x$-part of the ordered pair is also called the “run” (because you run back and forth), and the $ y$-part is the “rise” (because you rise/fall up and down). 
As an example, if we were to graph the situation where one makes $10 an hour at the mall, it turns out that the graph will be a straight line (linear), and every point on the line will give the $ x$–to-$ y$ (hours-to-earnings) solution. We can create a “T-chart” (do you see the “T”?) and then graph the line. Here is the equation, T-chart, and graph:
| T-chart | Graph | ||||||||||||
| $ \displaystyle y=10x$
|  |
In other words, when we graph the line, we can go over (back and forth) to see what the hours are and then look up to see how much we would make with that many hours. You can think of the $ x$ as the “question” on the bottom where you go back and forth, and then look up and down to get the “answer” where the $ y$ is – the answers are all on the line! The ordered pair $ (x,y)$ represents $ \left( {\text{number of hours you work, amount of money you make}} \right)$. We can just ignore the negative numbers (to the left of 0 and underneath 0); these are meaningless in this situation.
This concept is used every day for many different applications in the world! Of course, the relationship between $ x$ and $ y$ is not always linear (a straight line), but for now, we’ll work with linear relationships.
Any relation that has only one “answer” ($ y$) for each “question” ($ x$) is a function, which we’ll learn more about in the Algebraic Functions, including Domain and Range section.
Slope-Intercept Formula – the most “famous”
In the example above, the “$ y$-intercept”, or where the line hits the $ y$-axis is $ 0$ (this is actually at $ x=0$), and the coordinate point is $ \left( {0,0} \right)$. This is because when we work $ 0$ hours, we get paid $ 0$ dollars.
The most common equation for graphing a line is $ y=mx+b$, where $ m$ is the slope of the line, and $ b$ is the $ y$-intercept. This is called the slope-intercept formula for line, since it contains the slope and the $ y$-intercept. We really don’t know why they use “$ m$” for slope and “$ b$” for the $ y$-intercept; it’s one of those great mysteries in life!
The slope of the line is the rate of change; in our example, it is the rate per hour that you make working at the store. Think of the slope as how much you go up (or down) for going over (or back) one value of $ x$. Also, if you see words like “$ m$ for every (one) thing” or “$ m$ per (one) thing”, $ m$ would be the slope). In our case, when we go over one on the bottom (for example, from $ 2$ to $ 3$), we go up $ 10$ on the graph (from $ 20$ to $ 30$). It’s as easy as that! The slope can also be thought as the “rise” over the “run”. Again, don’t worry why; this is how someone just defined it. The $ y$-intercept is a “starting point” since it’s when $ x=0$.
Here’s another “real world” example. At the state far, it costs $ $4$ to get in (entrance fee), and then $ $2$ for every ride. The amount you spend ($ y$) depends on how many rides you take ($ x$); in other words, $ y=2x+4$. This can be interpreted as “The amount you spend at the fair is equal to 2 times the number of rides you take, plus the $4 to get in.” Here is a T-chart and graph:
| T-chart | Graph | ||||||||||
| $ \displaystyle y=2x+4$
|  |
Where does the line cross the $ y$-axis, in other words, where on the line is $ x=0$? This is before we’ve been on any rides. This is at $ y=4$; this is the $ y$-intercept. This is our amount to get into the fair (say, the “starting point”). The actual point is $ (0,4)$ (go back and forth $ 0$, go up $ 4$).
Look at the slope (or how much the line is slanted up, in this case). Do you see that for every time the graph goes up $ 2$ (“rise”), it goes over $ 1$ (“run”)? Our slope represents how fast we are spending money on the rides, but has nothing to do with the entrance fee. Note that you have to be really careful to compare slopes of different graphs, since the scales (how far apart the $ x$’s are, and the $ y$’s are) may be a little different, like in our graph.
Mathematically, to get the slope you can just take any two points and subtract the $ y$-values from each other and put that in the numerator, and then subtract the $ x$-values from each other (in same order that you subtracted the $ y$-values) and put that in the denominator. (Note that the little 1’s and 2’s are called subscripts, since they are a little bit below the actual letter. They just mean like the first $ x$, first $ y$ and then the second $ x$, second $ y$, and so on.) Here’s the formula: $ \displaystyle \text{slope}=\frac{{\text{rise}}}{{\text{run}}}=\frac{{({{y}_{2}}-{{y}_{1}})}}{{({{x}_{2}}-{{x}_{1}})}}$, where $ \left( {{{x}_{1}},{{y}_{1}}} \right)$ and $ \left( {{{x}_{2}},{{y}_{2}}} \right)$ are points on the line. Here is an example using points $ (2,8)$ and $ (6,16)$: $ \displaystyle \text{slope}=\frac{{\text{rise}}}{{\text{run}}}=\frac{{({{y}_{2}}-{{y}_{1}})}}{{({{x}_{2}}-{{x}_{1}})}}\,=\,\frac{{16-8}}{{6-2}}=\frac{8}{4}=\frac{2}{1}=2$. It doesn’t matter which point is first, as long as both the top and bottom are started with the same point.
Note that positive slopes go the right and up (like you’re throwing a ball if you’re right-handed), and negative slopes go to the right and down.
Here’s an example on how to graph from a slope-intercept formula, using another equation:
| Slope-Intercept Formula | Graph |
| $ \displaystyle y=-\frac{2}{3}x-2$ First graph the point $ (0,-2)$, since the $ y$-intercept is $ -2$. When $ y$ is negative, it’s below the $ x$-axis.
Then graph one more point using the slope, which is $ \displaystyle -\frac{2}{3}$. Since the slope is negative, go either to the right and down, or to the left and up. (One of the parts of the slope must be negative).
The numerator contains the “rise” (up or down) and the denominator the “run” (left and right). Note that if the slope is an integer, the “run” is $ 1$. |  |
Note that if we were given a linear equation in another form, such as $ \displaystyle x=7y+3$, we could do some algebra to get it to slope-intercept form: $ \require{cancel} \displaystyle x=7y+3;\,x-3=7y;\,\frac{{x-3}}{7}=\frac{{\cancel{7}y}}{{\cancel{7}}};\,y=\frac{x}{7}-\frac{3}{7}$. This line has a slope of $ \displaystyle \frac{1}{7}$ and a $ y$-intercept of $ \displaystyle -\frac{3}{7}$.
Point-Slope Formula
There is another formula that’s commonly used to graph a line, specifically when you have a slope and a random point that’s not the $ y$-intercept. This is called the point-slope formula, since it contains the slope and that random point: $ \displaystyle y-{{y}_{1}}=\,m\left( {x-{{x}_{1}}} \right)$, where $ \left( {{{x}_{1}},{{y}_{1}}} \right)$ is any point on the line. And we can always take this formula and put it back into the slope-intercept line or standard equation by doing a little bit of algebra!
Let’s say we know $ (2,1)$ is on a line, and the slope is $ -3$. Then we have an equation for the line: $ y-1=-3\left( {x-2} \right)$. (To put this form into the slope-intercept form, distribute the $ -3$ and isolate the $ y$ on the left: $ y-1=-3\left( {x-2} \right);\,\,y-1=-3x+6;\,\,y=-3x+7\,$.)
Here’s an example on how to graph from a point-slope formula:
| Point-Slope Formula | Graph |
| $ y-1=-3\left( {x-2} \right)$
First graph the point $ (2,1)$. Remember that the point-slope equation has minus signs, so $ x-2$ gives us $ 2$ for $ x$, and $ y-1$ will gives us $ 1$ for $ y$.
Since the slope is $ -3$, go up $ 3$ and back $ 1$, or down $ 3$ and forward $ 1$. |  |
Standard Form/Intercepts Method
The Standard Form of a linear equation is supposedly the “most elegant” form of a line and is used most often for graphing using the Intercepts Method, as shown below. It is in the form $ Ax+By=C$, where $ A$, $ B$, and $ C$ are integers and $ A$ is positive. (You may see the similar general form of a linear equation, which is $ Ax+By-C=0$.) We can “turn” the slope-intercept form or any other form of a linear equation into the standard form with a little bit of algebra. For example, $ \displaystyle y=-\frac{2}{3}x-2;\,\,\,3y=-2x-6;\,\,2x+3y=-6\,$.
To graph the standard form using the intercepts method (which I also call the “cover up” method), you obtain the $ x$- and $ y$-intercepts by “covering up” the other variable and its coefficient. This is because the $ \boldsymbol{y}$–intercept occurs when $ \boldsymbol{x=0}$, and the $ \boldsymbol{x}$–intercept occurs when $ \boldsymbol{y=0}$ (look at a graph, and you might be able to see this!).
Here’s an example on how to graph from the Standard formula:
| Intercepts/”Cover up” Method | Graph |
| $ 3x+4y=12$
“Cover up” the $ 4y$ to get $ 3x=12;\,\,x=4$ , and “cover up” the $ 3x$ to get $ 4y=12;\,\,y=3$).
Graph our two points, $ \left( {4,0} \right)$ ($ x$-intercept) and $ \left( {0,3} \right)$ ($ y$-intercept). When we do the “cover up” method, we’ll have one point on each of the axes. |  |
You can use this method for any equation in the form $ \boldsymbol {Ax+By=C}$, but it’s much easier when $ A$ and $ B$ go into $ C$ (otherwise you are dealing with fractions). Do you see how the $ x$–intercept is $ \displaystyle \left( {\frac{C}{A},\,0} \right)$, and the $ y$–intercept is $ \displaystyle \left( {0,\,\frac{C}{B}} \right)$? Pretty cool, isn’t it?
Horizontal and Vertical Lines
Horizontal lines are when “$ y=$ a number”; for example, “$ y=5$”. There is no $ x$ in the equation, so this is a horizontal line where $ y$ is always $ 5$, no matter what $ x$ is. The slope is zero (0), since $ 0x$ means there is no $ x$. You can remember this since a “$ y$” looks like an upside down “$ h$”, and “$ h$” is the beginning of the word “horizontal”. The line “$ y=0$” sits on the $ x$-axis.
Vertical lines are when “$ x=$ a number”; for example, “$ x=2$”. There is no $ y$ in the equation, so this is a vertical line where $ x$ always equals $ 2$, no matter what $ y$ is. The slope is undefined, like the line is falling from the sky. You can remember this since you can draw a “$ v$” in an “$ x$”, and “$ v$” is the beginning of the word “vertical”. The line “$ x=0$” sits on the $ y$-axis.
Here are some examples:
 “0” Slope (horizontal line) |  Undefined Slope (vertical line) |
| There is no “$ x$” in the equation, since it doesn’t matter what “$ x$ is (“$ y$ always stays the same). Example: At an all-you-can-eat restaurant, you pay the same amount (say $ \$5$), no matter how many plates of food you eat (“$ x$”). | There is no “$ y$” in the equation, since it doesn’t matter what $ y$ is ($ x$ is always stays the same). Example: Think of an elevator going up and down where $ x=2$. We’ll learn later that this really doesn’t make sense, since every $ x$-value should really only have one $ y$-value to be a function. |
Parallel and Perpendicular Lines
Parallel Lines
Parallel lines are lines that look like railroad tracks or the two lines in an equal sign; they never cross each other. Parallel lines have the same slope. Do you see how this is true If you look at any two different points on the lines, you would go back and forth, and up and down the same amount. You may also see parallel as $ \parallel $.
Here’s an example: you and your friend Madison make the same amount of money (say $ \$10$ an hour) at the mall, but your friend got a bonus of a $ \$25$ gift certificate for the store when she started working. If you include the $ \$25$ that she got when she started working, if the two of you work the same number of hours, you will never catch up with her. Here are the two equations on a graph: 
Note that special cases of parallel lines are two horizontal lines (such as $ y=3$ and $ y=4$) and two vertical lines (such as $ x=-1$ and $ x=4$).
Perpendicular Lines
Perpendicular lines are lines that cross each other exactly like the corner of a piece of paper; they meet at what we call a “right” angle, or (if you’ve had Geometry), a 90 degree angle. You may also see perpendicular as $ \bot $.
There’s something very peculiar about the slopes of perpendicular lines; their slopes are what we call negative reciprocals or opposite reciprocals of each other. This sounds fancy, but all it means is that we take the first slope, change the sign (make it positive if it’s negative, or make it negative if it’s positive) and then flip it (for example, $ 2$ would flip to be $ \displaystyle \frac{1}{2}$, $ \displaystyle \frac{2}{3}$ would flip to be $ \displaystyle \frac{3}{2}$) – we will have the slope of the line perpendicular to it. It makes sense that we take the negative reciprocal, if you think about it. Since the lines crisscross each other, it makes sense that if one slope is positive, the other is negative (or if one has zero slope, the other would have undefined slope). If you start at a point and go up $ 2$ and over $ 1$, for example, it would make sense that if you go over $ 2$ and down$ 1$, you’ll end up with a perfect corner: a right angle. Try a few on a piece of graph paper. 
Here are some combinations of slopes that are perpendicular: $ \displaystyle \frac{2}{3}$ and $ \displaystyle -\frac{3}{2}$, $ 4$ and $ \displaystyle -\frac{1}{4}$, $ 1$ and $ -1$, $ \displaystyle -\frac{5}{8}$ and $ \displaystyle \frac{8}{5}$, $ 0$ (for example, horizontal line $ y=3$) and UNDEFINED (for example, vertical line $ x=2$, since we can’t divide by $ 0$!).
Here are the types of parallel and perpendicular line problems you might see:
| Parallel/Perpendicular Problem | Solution |
| Find the equation of the line (in slope-intercept form) that goes through the point $ (2,1)$ and is parallel to the line $ y=4x+3$. | We want an equation with slope $ 4$, since it has to be parallel to the given equation. The $ y$-intercept will be different, and we need to solve for that using the given point. Use the slope-intercept equation; to solve for $ b$, plugging in $ 2$ for $ x$ and $ 1$ for $ y$ for point $ (2,1)$: $ \displaystyle \begin{array}{c}y=mx+b;\,\,y=4x+b\\1=4(2)+b;\,\,1=\,8+b;\,\,b=-7\\y=\,4x-7\end{array}$ |
| Find the equation of the line (in slope-intercept, point-slope, and standard form) that goes through the point $ (2,1)$ and is perpendicular to the line $ y=4x+3$. | The slope of a line perpendicular is $ \displaystyle -\frac{1}{4}$, since this is the negative reciprocal of $ 4$. To get the point-slope form of the new equation, use the given point $ (2,1)$: $ \displaystyle y-{{y}_{1}}=m\left( {x-{{x}_{1}}} \right);\,\,y-1=-\frac{1}{4}\left( {x-2} \right)$ Now use algebra to get the slope-intercept form: $ \displaystyle y=-\frac{1}{4}\left( {x-2} \right)+1;\,\,\,\,y=-\frac{1}{4}x+\frac{1}{2}+1;\,\,\,\,y=-\frac{1}{4}x+\frac{3}{2}$ Use more algebra to get the Standard form: $ \displaystyle y=-\frac{1}{4}x+\frac{3}{2};\,\,4y=-x+6;\,\,x+4y=6$ |
Obtaining an Equation for a Line
Sometimes you’ll want to get the equation for line, given that you have two points on that line. Remember that two points make any line; try this by drawing two points on a piece of paper and trying to draw more than one line between them; you can’t! Here’s an example:
Suppose, with an example from another fair in another state, you know that Jane rode $ 4$ rides and spent $ \$24$ and Judy rode $ 2$ rides and only spent $ \$16$. From these two points, you want a general equation of the line, to see how much it would cost, say, if you were to ride $ 7$ rides.
We’ve learned that we can use the slope-intercept equation $ \boldsymbol{y=mx+b}$, or the slope-point equation $ \boldsymbol{{y-{{y}_{1}}=m\left( {x-{{x}_{1}}} \right)}}$; we can use either equation. For both equations, we first need to find the slope, which is the change in $ y$’s over change in $ x$’s. Since the $ x$ is the number of rides, and the $ y$ is the price, Jane’s point on the line would be $ (4,24)$, and Judy’s would be $ (2,16)$. Remember that we have to start with the same point in the subtraction on the top and the bottom: $ \displaystyle \text{slope}=\frac{{\text{rise}}}{{\text{run}}}\,=\,\frac{{({{y}_{2}}-{{y}_{1}})}}{{({{x}_{2}}-{{x}_{1}})}}\,=\,\frac{{24-16}}{{4-2}}\,=\,\frac{8}{2}\,=4$. Thus, each ride costs $ \$4$, since the slope is a rate.
At this point, it’d be easier to use the point-slope equation, but you could alternatively use the slope-intercept equation. You can use either point; it will still get to the same equation! You can also see how we can get the slope-intercept form of the equation with a little algebra. (After getting the equation, we can see that to ride $ 7$ rides, the cost would be $ y=4\left( 7 \right)+8=\$36$.)
| Point-Slope Point $ (4,24)$: | Point-Slope: Point $ (2,16)$: | Slope-Intercept: Point $ (4,24)$: | Slope-Intercept Point $ (2,16)$: |
| $ \displaystyle \begin{array}{l}y-{{y}_{1}}=\,\,m\left( {x-{{x}_{1}}} \right)\\\\y-24=4(x-4)\\\underline{\begin{array}{l}y-24=4x-16\\\,\,\,\,\,+24=\,\,\,\,\,\,\,+\,\,24\end{array}}\\\,y\,\,\,\,\,\,\,\,\,\,\,\,=4x+8\,\,\,\,\,\,\,\surd \end{array}$ | $ \displaystyle \begin{array}{l}y-{{y}_{1}}=\,m\left( {x-{{x}_{1}}} \right)\\\\y-16=4(x-2)\\\underline{\begin{array}{l}y-16=4x-8\\\,\,\,\,\,+16=\,\,\,\,\,\,+\,\,16\end{array}}\\\,y\,\,\,\,\,\,\,\,\,\,\,\,\,=4x+8\,\,\,\,\,\,\,\surd \end{array}$ | $ \displaystyle \begin{array}{l}\,\,\,\,\,\,y\,\,=\,mx+b\\\\\,\,\,\,\,\,\,y=\,4x+b\\\,\,\,\,24=4(4)+b\\\,\,\,\,24=16+b\\\underline{{-16=\,-16}}\\\,\,\,\,\,\,8\,=\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,b\\y\,=\,4x+8\,\,\,\,\,\,\,\surd \end{array}$ | $ \displaystyle \begin{array}{l}\,\,\,\,\,y=mx+b\\\\\,\,\,\,\,y=4x+b\\\,\,\,16=4(2)+b\\\,\,\,16=8+b\\\,\underline{{-8\,=\,-8}}\\\,\,\,\,\,8\,=\,\,\,\,\,\,\,\,\,\,b\\y\,=\,4x+8\,\,\,\,\,\,\,\surd \end{array}$ |
To graph this equation, we can either make a T-chart like the one below and plot points (remember, you actually only need 2 points), or plot the $ y$-intercept $ (0,8)$ and then use the slope to go up $ 4$, over $ 1$, and then draw the line:
| Linear Function T-chart | Graph | ||||||||||
|  |
Again, do you see how when we make coordinate graphs, we look for the “question” on the bottom (the “$ x$”) and then go over and find the “answer” on the line (the “$ y$” part of the line)? This is a very important concept.
Distance and Midpoint Formulas
There are a couple of formulas that you’ll learn (either in Geometry or Algebra or both) that have to do with the Coordinate System and points; these are the Distance Formula and the Midpoint Formula.
Distance Formula
The Distance Formula is exactly what is says: it’s the distance between two given points. It gives you a single number that is the distance measured between two points, say $ \left( {{{x}_{1}},{{y}_{1}}} \right)$ and $ \left( {{{x}_{2}},{{y}_{2}}} \right)$: $ \text{Distance}=\sqrt{{{{{\left( {{{x}_{2}}-{{x}_{1}}} \right)}}^{2}}+{{{\left( {{{y}_{2}}-{{y}_{1}}} \right)}}^{2}}}}$. It doesn’t matter which point is $ \left( {{{x}_{1}},{{y}_{1}}} \right)$ and which point is $ \left( {{{x}_{2}},{{y}_{2}}} \right)$, as long as you always start with the same point when you do the subtracting. Here are some examples:
| Distance Formula Problem | Solution |
| Find the distance between the points $ (2,–3)$ and $ (7,2)$. | Let $ \left( {{{x}_{1}},{{y}_{1}}} \right)$ be $ (2,-3)$ and $ \left( {{{x}_{2}},{{y}_{2}}} \right)$ be $ (7,2)$. The distance between the two points is: $ \sqrt{{{{{\left( {{{x}_{2}}-{{x}_{1}}} \right)}}^{2}}+{{{\left( {{{y}_{2}}-{{y}_{1}}} \right)}}^{2}}}}=\sqrt{{{{{\left( {7-2} \right)}}^{2}}+{{{\left( {2-\left( {-3} \right)} \right)}}^{2}}}}=\sqrt{{50}}\approx 7.071$. Notice in the graph that the Distance Formula uses the Pythagorean Theorem, if you were to draw a right triangle “between” the points: |
| The distance between points $ \left( {-1,5} \right)$ and $ \left( {5,{{y}_{2}}} \right)$ is $ 10$. What are the two possible values for $ \displaystyle {{y}_{2}}$? | Set the distance between points $ \left( {-1,5} \right)$ (say $ \left( {{{x}_{1}},{{y}_{1}}} \right)$), and $ \left( {5,{{y}_{2}}} \right)$ to $ 10$, and solve for $ \displaystyle {{y}_{2}}$; don’t forget to include both the positive and negative value when taking the square root: $ \displaystyle \begin{array}{c}\sqrt{{{{{\left( {{{x}_{2}}-{{x}_{1}}} \right)}}^{2}}+{{{\left( {{{y}_{2}}-{{y}_{1}}} \right)}}^{2}}}}=\sqrt{{{{{\left( {5-\left( {-1} \right)} \right)}}^{2}}+{{{\left( {{{y}_{2}}-5} \right)}}^{2}}}}=10\\\sqrt{{{{6}^{2}}+{{{\left( {{{y}_{2}}-5} \right)}}^{2}}}}=10;\,\,\,\,36+{{\left( {{{y}_{2}}-5} \right)}^{2}}=100\\{{\left( {{{y}_{2}}-5} \right)}^{2}}=64;\,\,\,{{y}_{2}}-5=\pm 8;\,\,\,\,{{y}_{2}}=-3,13\end{array}$ |
Midpoint Formula
The Midpoint Formula gives you the point that is exactly half-way between two given points, say $ \left( {{{x}_{1}},{{y}_{1}}} \right)$ and $ \left( {{{x}_{2}},{{y}_{2}}} \right)$: $ \displaystyle \text{Midpoint}=\left( {\frac{{{{x}_{1}}+{{x}_{2}}}}{2},\frac{{{{y}_{1}}+{{y}_{2}}}}{2}} \right)$. Note that the result is an ordered pair and not just a number (thus, midpoint). I like to think of the midpoint formula as just the averages of the two points. Here are some examples:
| Midpoint Formula Problem | Solution |
| Find the midpoint between the points $ (2,-3)$ and $ (7,2)$. | Let $ \left( {{{x}_{1}},{{y}_{1}}} \right)$ be $ (2,-3)$ and $ \left( {{{x}_{2}},{{y}_{2}}} \right)$ be $ (7,2)$. The midpoint between the two points is: $ \displaystyle \left( {\frac{{{{x}_{1}}+{{x}_{2}}}}{2},\,\frac{{{{y}_{1}}+{{y}_{2}}}}{2}} \right)=\left( {\frac{{2+7}}{2},\,\frac{{-3+2}}{2}} \right)=\left( {\frac{9}{2},\,\frac{{-1}}{2}} \right)=\left( {\frac{9}{2},\,-\frac{1}{2}} \right)$. You can see that this point appears to be exactly in the middle of the two points (along the diagonal). |
| $ (5,0)$ is the midpoint of $ (3,4)$ and another point. Find the other point. | We still need to use the midpoint formula, but have to solve backwards to get the other point: $ \displaystyle \left( {\frac{{3+{{x}_{2}}}}{2},\,\frac{{4+{{y}_{2}}}}{2}} \right)=\left( {5,0} \right);\,\,\,\,\frac{{3+{{x}_{2}}}}{2}=5\,\,\text{and}\,\,\frac{{4+{{y}_{2}}}}{2}=0$. Solve for $ {{x}_{2}}$ and $ {{y}_{2}}$ to get $ 7$ and $ -4$, respectively. Thus, the other point is $ (7,-4)$. You can always check your answer back by finding the midpoint of $ (3,4)$, and $ (7,-4)$; you get $ (5,0)$! |
Linear Inequalities in Two Variables
If we were to graph two or more non-parallel linear equations, the point that they intersect is the solution to their system of equations; we will see how to do this in here in the Systems of Equations section. We can also graph inequalities with two variables or more to solve them; they will typically have an infinite number of solutions. These types of graphs are actually kind of pretty and fun to do – almost like artwork! Just remember that the areas shaded are where the equation “works”; in other words, where the solutions are.
When graphing the equations in the slope-intercept method (“$ \displaystyle y=mx+b$”) formula (or another method, like the intercept method), we look which way the inequality sign is, with respect to the positive coefficient of $ \boldsymbol {y}$. When we have “$ y<$”, we always shade in under the line that we draw (or to the left, if we have a vertical line). Think of “less than” as “raining down” from the graph. Note that we need a positive coefficient of $ y$ for this to work. When we have “$ y>$”, we always shade above the line that we draw (or to the right, if we have a vertical line). Think of “greater than” as “raining up” from the graph. (I know – it doesn’t really “rain up”, but I still like to explain the graphs that way.) Note that we need a positive coefficient of $ y$ for this to work. You can always plug in an $ (x,y)$ ordered pair to see if it shows up in the shaded areas (which means it’s a solution), or the unshaded areas (which means it’s not a solution.) For an example of this, see the first inequality below.
With “$ <$” and “$ >$” inequalities, draw a dashed (or dotted) line to indicate that we’re not including that line (exclusive), whereas with “$ \le$” and “$ \ge$”, draw a regular line, to indicate that we are including it in the solution (inclusive). To remember this, I think about the fact that “$ <$” and “$ >$” have less pencil marks than “$ \le$” and “$ \ge$”, so there is less pencil used when you draw the lines on the graph. You can also remember this by thinking the line under the “$ \le$” and “$ \ge$” means you draw a solid line on the graph.
The last example is a “Compound Inequality” since it involves more than one inequality. The solution set is the ordered pairs that satisfy both inequalities; it is indicated by the darker shading
We’ll use these types of graphs when we work in the Introduction to Linear Programming section later.
Here are some examples:
| Inequality/Explanation | Graph |
| $ y<2x+4$
First, graph the line $ y=2x+4$. For a “$ y<$” inequality, shade under the graph, since it “rains down”. Use a dashed line since there is a “$ <$ ”.
Plug in $ \left( {5,0} \right)$ to see if it shows up as a solution. $ \left( {5,0} \right)$ is in the shaded area, so that it satisfies the inequality: $ \displaystyle 0<2\left( 5 \right)+4\,\,\,\,\,\surd $. |  |
| $ 3x-4y\le 12$
The easiest way to graph this is with the “cover up” or intercepts method, since there are variables on one side and a constant on the other.
To graph using the intercepts method, cover up the $ 4y$ (making $ y=0$) and the $ x$-intercept is 4, since $ 3\times 4=12$. Then cover up the $ 3x$ (making $ x=0$) and the $ y$-intercept is –3, since $ -4\times -3=12$.
Be careful though since the $ -4y$ is before the $ \le $ sign. We need a positive $ y$, so the $ -4y$ would end up on the side of the “greater than”. Thus, shade above the line (“rains up”). Note that there is a solid line because of the “$ \ge $”. |  |
| $ x>3$
For a vertical line, with “greater than”, shade to the right. (With “less than”, $ <$, shade to the left.)
Note: For a horizontal line (like $ y>3$), shade above the line, and with $ y<3$, shade below the line. |  |
| Compound Inequality
$ y<2x+4$ and $ y\le 5$
Draw both lines and fill in the inequalities separately.
The solution set is the ordered pairs that satisfy both inequalities; it is indicated by the darker shading. Note that part of this solution would be inclusive (the solid lines) and part would be exclusive (the dotted line). |  |
Learn these rules, and practice, practice, practice!
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On to Direct, Inverse, Joint and Combined Variation – you are ready!