2.1 The Rectangular Coordinate Systems and Graphs
Figure 1
Tracie set out from Elmhurst, IL, to go to Franklin Park. On the way, she made a few stops to do errands. Each stop is indicated by a red dot in Figure 1. Laying a rectangular coordinate grid over the map, we can see that each stop aligns with an intersection of grid lines. In this section, we will learn how to use grid lines to describe locations and changes in locations.
2.1.1 Plotting Ordered Pairs in the Cartesian Coordinate System
An old story describes how seventeenth-century philosopher/mathematician René Descartes, while sick in bed, invented the system that has become the foundation of algebra. According to the story, Descartes was staring at a fly crawling on the ceiling when he realized that he could describe the fly’s location in relation to the perpendicular lines formed by the adjacent walls of his room. He viewed the perpendicular lines as horizontal and vertical axes. Further, by dividing each axis into equal unit lengths, Descartes saw that it was possible to locate any object in a two-dimensional plane using just two numbers—the displacement from the horizontal axis and the displacement from the vertical axis.
While there is evidence that ideas similar to Descartes’ grid system existed centuries earlier, it was Descartes who introduced the components that comprise the Cartesian coordinate system, a grid system having perpendicular axes. Descartes named the horizontal axis the \(\mathbf{x}\)-axis and the vertical axis the \(\mathbf{y}\)-axis.
The Cartesian coordinate system, also called the rectangular coordinate system, is based on a two-dimensional plane consisting of the \(x\)-axis and the \(y\)-axis. Perpendicular to each other, the axes divide the plane into four sections. Each section is called a quadrant; the quadrants are numbered counterclockwise as shown in Figure 2
Figure 2
The center of the plane is the point at which the two axes cross. It is known as the origin, or point \((0,0).\) From the origin, each axis is further divided into equal units: increasing, positive numbers to the right on the \(x\)-axis and up the \(y\)-axis; decreasing, negative numbers to the left on the \(x\)-axis and down the \(y\)-axis. The axes extend to positive and negative infinity as shown by the arrowheads in Figure 3.
Figure 3
Each point in the plane is identified by its \(\mathbf{x}\)-coordinate, or horizontal displacement from the origin, and its \(\mathbf{y}\)-coordinate, or vertical displacement from the origin. Together, we write them as an ordered pair indicating the combined distance from the origin in the form \(\left( {x,y} \right).\) An ordered pair is also known as a coordinate pair because it consists of \(x\)- and \(y\)-coordinates. For example, we can represent the point \((3,-1)\) in the plane by moving three units to the right of the origin in the horizontal direction, and one unit down in the vertical direction. See Figure 4.
Figure 4
When dividing the axes into equally spaced increments, note that the \(x\)-axis may be considered separately from the \(y\)-axis. In other words, while the \(x\)-axis may be divided and labeled according to consecutive integers, the \(y\)-axis may be divided and labeled by increments of 2, or 10, or 100. In fact, the axes may represent other units, such as years against the balance in a savings account, or quantity against cost, and so on. Consider the rectangular coordinate system primarily as a method for showing the relationship between two quantities.
Plot the points \(\left( {-2,4} \right),\) \(\left( {3,3} \right),\) and \(\left( {0,-3} \right)\) in the plane.
Solution (click to reveal)
To plot the point \(\left( {-2,4} \right),\) begin at the origin. The \(x\)-coordinate is –2, so move two units to the left. The \(y\)-coordinate is 4, so then move four units up in the positive y direction.
To plot the point \(\left( {3,3} \right),\) begin again at the origin. The \(x\)-coordinate is 3, so move three units to the right. The \(y\)-coordinate is also 3, so move three units up in the positive y direction.
To plot the point \(\left( {0,-3} \right),\) begin again at the origin. The \(x\)-coordinate is 0. This tells us not to move in either direction along the \(x\)-axis. The \(y\)-coordinate is –3, so move three units down in the negative \(y\) direction. See the graph in Figure 5.
Figure 5
Note that when either coordinate is zero, the point must be on an axis. If the \(x\)-coordinate is zero, the point is on the \(y\)-axis. If the \(y\)-coordinate is zero, the point is on the \(x\)-axis.
2.1.2 Graphing Equations by Plotting Points
We can plot a set of points to represent an equation. When such an equation contains both an x variable and a y variable, it is called an equation in two variables. Its graph is called a graph in two variables. Any graph on a two-dimensional plane is a graph in two variables.
Suppose we want to graph the equation \(y = 2x - 1.\) We can begin by substituting a value for \(x\) into the equation and determining the resulting value of \(y\). Each pair of \(x\)- and \(y\)-values is an ordered pair that can be plotted. Table 1 lists values of \(x\) from –3 to 3 and the resulting values for \(y\).
| \(x\) | \(y = 2x - 1\) | \(\left( {x,y} \right)\) |
| \(-3\) | \(y = 2(-3) - 1 = -7\) | \(\left( {-3,-7} \right)\) |
| \(-2\) | \(y = 2(-2) - 1 = -5\) | \(\left( {-2,-5} \right)\) |
| \(-1\) | \(y = 2(-1) - 1 = -3\) | \(\left( {-1,-3} \right)\) |
| \(0\) | \(y = 2(0) - 1 = -1\) | \(\left( {0,-1} \right)\) |
| \(1\) | \(y = 2(1) - 1 = 1\) | \(\left( {1,1} \right)\) | |
| \(2\) | \(y = 2(2) - 1 = 3\) | \(\left( {2,3} \right)\) | |
| \(3\) | \(y = 2(3) - 1 = 5\) | \(\left( {3,5} \right)\) | |
Table 1
We can plot the points in the table. The points for this particular equation form a line, so we can connect them. See Figure 6. This is not true for all equations.
Figure 6
Note that the \(x\)-values chosen are arbitrary, regardless of the type of equation we are graphing. Of course, some situations may require particular values of \(x\) to be plotted in order to see a particular result. Otherwise, it is logical to choose values that can be calculated easily, and it is always a good idea to choose values that are both negative and positive. There is no rule dictating how many points to plot, although we need at least two to graph a line. Keep in mind, however, that the more points we plot, the more accurately we can sketch the graph.
Graph the equation \(y = - x + 2\) by plotting points.
Solution (click to reveal)
First, we construct a table similar to Table 2. Choose \(x\) values and calculate \(y\).
| \(x\) | \(y = - x + 2\) | \(\left( {x,y} \right)\) |
| \(-5\) | \(y = - (-5) + 2 = 7\) | \(\left( {-5,7} \right)\) | |
| \(-3\) | \(y = - (-3) + 2 = 5\) | \(\left( {-3,5} \right)\) | |
| \(-1\) | \(y = - (-1) + 2 = 3\) | \(\left( {-1,3} \right)\) | |
| \(0\) | \(y = - (0) + 2 = 2\) | \(\left( {0,2} \right)\) | |
| \(1\) | \(y = - (1) + 2 = 1\) | \(\left( {1,1} \right)\) | |
| \(3\) | \(y = - (3) + 2 = -1\) | \(\left( {3,-1} \right)\) |
| \(5\) | \(y = - (5) + 2 = -3\) | \(\left( {5,-3} \right)\) |
Table 2
Now, plot the points. Connect them if they form a line. See Figure 7
Figure 7
Construct a table and graph the equation by plotting points: \(y = \frac{1}{2}x + 2.\)
Solution (click to reveal)
| \(x\) | \(y = \frac{1}{2}x + 2\) | \(\left( {x,y} \right)\) |
| \(-2\) | \(y = \frac{1}{2}(-2) + 2 = 1\) | \(\left( {-2,1} \right)\) | |
| \(-1\) | \(y = \frac{1}{2}(-1) + 2 = \frac{3}{2}\) | \(\left( {- 1,\frac{3}{2}} \right)\) |
| \(0\) | \(y = \frac{1}{2}(0) + 2 = 2\) | \(\left( {0,2} \right)\) | |
| \(1\) | \(y = \frac{1}{2}(1) + 2 = \frac{5}{2}\) | \(\left( {1,\frac{5}{2}} \right)\) |
| \(2\) | \(y = \frac{1}{2}(2) + 2 = 3\) | \(\left( {2,3} \right)\) | |
Table 3
2.1.3 Graphing Equations with a Graphing Utility
Most graphing calculators require similar techniques to graph an equation. The equations sometimes have to be manipulated so they are written in the style \(y = \operatorname{\_\_\_\_\_}.\) The TI-84 Plus, and many other calculator makes and models, have a mode function, which allows the window (the screen for viewing the graph) to be altered so the pertinent parts of a graph can be seen.
For example, the equation \(y = 2x - 20\) has been entered in the TI-84 Plus shown in Figure 8a. In Figure 8b, the resulting graph is shown. Notice that we cannot see on the screen where the graph crosses the axes. The standard window screen on the TI-84 Plus shows \(-10 \leq x \leq 10,\) and \(-10 \leq y \leq 10.\) See Figure 8c.
Figure 8 a. Enter the equation. b. This is the graph in the original window. c. These are the original settings.
By changing the window to show more of the positive \(x\)-axis and more of the negative \(y\)-axis, we have a much better view of the graph and the \(x\)- and \(y\)-intercepts. See Figure 9a and Figure 9b.
Figure 9 a. This screen shows the new window settings. b. We can clearly view the intercepts in the new window.
Use a graphing utility to graph the equation: \(y = - \frac{2}{3}x + \frac{4}{3}.\)
Solution (click to reveal)
Enter the equation in the y= function of the calculator. Set the window settings so that both the \(x\)- and \(y\)- intercepts are showing in the window. See Figure 10.
Figure 10
2.1.4 Finding \(x\)-intercepts and \(y\)-intercepts
The intercepts of a graph are points at which the graph crosses the axes. The \(\mathbf{x}\)-intercept is the point at which the graph crosses the \(x\)-axis. At this point, the \(y\)-coordinate is zero. The \(\mathbf{y}\)-intercept is the point at which the graph crosses the \(y\)-axis. At this point, the \(x\)-coordinate is zero.
To determine the \(x\)-intercept, we set y equal to zero and solve for \(x\). Similarly, to determine the \(y\)-intercept, we set x equal to zero and solve for \(y\). For example, lets find the intercepts of the equation \(y = 3x - 1.\)
To find the \(x\)-intercept, set \(y = 0.\)
\[\begin{array}{ll} {\mspace{9mu} y = 3x - 1} & \\ {\mspace{9mu} 0 = 3x - 1} & \\ {\mspace{9mu} 1 = 3x} & \\ {\frac{1}{3} = x} & \\ \left( {\frac{1}{3},0} \right) & {x\text{−intercept}} \end{array}\]
To find the \(y\)-intercept, set \(x = 0.\)
\[\begin{array}{l} {y = 3x - 1} \\ {y = 3(0) - 1} \\ {y = -1} \\ {(0,-1)\mspace{54mu} y\text{−intercept}} \end{array}\]
We can confirm that our results make sense by observing a graph of the equation as in Figure 11. Notice that the graph crosses the axes where we predicted it would.
Figure 11
Find the intercepts of the equation \(y = -3x - 4.\) Then sketch the graph using only the intercepts.
Solution (click to reveal)
Set \(y = 0\) to find the \(x\)-intercept.
\[\begin{array}{l} {\quad y = -3x - 4} \\ {\quad 0 = -3x - 4} \\ {\quad 4 = -3x} \\ {- \frac{4}{3} = x} \\ {\left( {- \frac{4}{3},0} \right)\mspace{54mu} x\text{−intercept}} \end{array}\]
Set \(x = 0\) to find the \(y\)-intercept.
\[\begin{array}{l} {y = -3x - 4} \\ {y = -3(0) - 4} \\ {y = -4} \\ {(0,-4)\mspace{63mu} y\text{−intercept}} \end{array}\]
Plot both points, and draw a line passing through them as in Figure 12.
Figure 12
Find the intercepts of the equation and sketch the graph: \(y = - \frac{3}{4}x + 3.\)
Solution (click to reveal)
\(x\)-intercept is \(\left( {4,0} \right);\) \(y\)-intercept is \(\left( {0,3} \right).\)
2.1.5 Using the Distance Formula
Derived from the Pythagorean Theorem, the distance formula is used to find the distance between two points in the plane. The Pythagorean Theorem, \(a^{2} + b^{2} = c^{2},\) is based on a right triangle where a and \(b\) are the lengths of the legs adjacent to the right angle, and \(c\) is the length of the hypotenuse. See Figure 13.
Figure 13
The relationship of sides \(\left| {x_{2} - x_{1}} \right|\) and \(\left| {y_{2} - y_{1}} \right|\) to side \(d\) is the same as that of sides a and b to side \(c\). We use the absolute value symbol to indicate that the length is a positive number because the absolute value of any number is positive. (For example, \(|-3| = 3.\) ) The symbols \(\left| {x_{2} - x_{1}} \right|\) and \(\left| {y_{2} - y_{1}} \right|\) indicate that the lengths of the sides of the triangle are positive. To find the length \(c\), take the square root of both sides of the Pythagorean Theorem.
\[c^{2} = a^{2} + b^{2}\rightarrow c = \sqrt{a^{2} + b^{2}}\]
It follows that the distance formula is given as
\[d^{2} = {(x_{2} - x_{1})}^{2} + {(y_{2} - y_{1})}^{2}\rightarrow d = \sqrt{{(x_{2} - x_{1})}^{2} + {(y_{2} - y_{1})}^{2}}\]
We do not have to use the absolute value symbols in this definition because any number squared is positive.
Find the distance between the points \(\left( {-3,-1} \right)\) and \(\left( {2,3} \right).\)
Solution (click to reveal)
Let us first look at the graph of the two points. Connect the points to form a right triangle as in Figure 14.
Figure 14
Then, calculate the length of d using the distance formula.
\[\begin{array}{l} \\ \begin{array}{l} {d = \sqrt{{(x_{2} - x_{1})}^{2} + {(y_{2} - y_{1})}^{2}}} \\ {d = \sqrt{{(2 - (-3))}^{2} + {(3 - (-1))}^{2}}} \\ { = \sqrt{{(5)}^{2} + {(4)}^{2}}} \\ { = \sqrt{25 + 16}} \\ { = \sqrt{41}} \end{array} \end{array}\]
Let’s return to the situation introduced at the beginning of this section.
Tracie set out from Elmhurst, IL, to go to Franklin Park. On the way, she made a few stops to do errands. Each stop is indicated by a red dot in Figure 1. Find the total distance that Tracie traveled. Compare this with the distance between her starting and final positions.
Solution (click to reveal)
The first thing we should do is identify ordered pairs to describe each position. If we set the starting position at the origin, we can identify each of the other points by counting units east (right) and north (up) on the grid. For example, the first stop is 1 block east and 1 block north, so it is at \(\left( {1,1} \right).\) The next stop is 5 blocks to the east, so it is at \(\left( {5,1} \right).\) After that, she traveled 3 blocks east and 2 blocks north to \(\left( {8,3} \right).\) Lastly, she traveled 4 blocks north to \(\left( {8,7} \right).\) We can label these points on the grid as in Figure 15.
Figure 15
Next, we can calculate the distance. Note that each grid unit represents 1,000 feet.
- From her starting location to her first stop at \(\left( {1,1} \right),\) Tracie might have driven north 1,000 feet and then east 1,000 feet, or vice versa. Either way, she drove 2,000 feet to her first stop.
- Her second stop is at \(\left( {5,1} \right).\) So from \(\left( {1,1} \right)\) to \(\left( {5,1} \right),\) Tracie drove east 4,000 feet.
- Her third stop is at \(\left( {8,3} \right).\) There are a number of routes from \(\left( {5,1} \right)\) to \(\left( {8,3} \right).\) Whatever route Tracie decided to use, the distance is the same, as there are no angular streets between the two points. Let’s say she drove east 3,000 feet and then north 2,000 feet for a total of 5,000 feet.
- Tracie’s final stop is at \(\left( {8,7} \right).\) This is a straight drive north from \(\left( {8,3} \right)\) for a total of 4,000 feet.
Next, we will add the distances listed in Table 4.
| From/To | Number of Feet Driven |
|---|---|
| \(\left( {0,0} \right)\) to \(\left( {1,1} \right)\) | 2,000 | | |
| \(\left( {1,1} \right)\) to \(\left( {5,1} \right)\) | 4,000 | | |
| \(\left( {5,1} \right)\) to \(\left( {8,3} \right)\) | 5,000 | | |
| \(\left( {8,3} \right)\) to \(\left( {8,7} \right)\) | 4,000 | | |
| Total | 15,000 |
Table 4
The total distance Tracie drove is 15,000 feet, or 2.84 miles. This is not, however, the actual distance between her starting and ending positions. To find this distance, we can use the distance formula between the points \(\left( {0,0} \right)\) and \(\left( {8,7} \right).\)
\[\begin{array}{l} {d = \sqrt{{(8 - 0)}^{2} + {(7 - 0)}^{2}}} \\ { = \sqrt{64 + 49}} \\ { = \sqrt{113}} \\ { \approx 10.63\mspace{9mu}\text{units}} \end{array}\]
At 1,000 feet per grid unit, the distance between Elmhurst, IL, to Franklin Park is 10,630.14 feet, or 2.01 miles. The distance formula results in a shorter calculation because it is based on the hypotenuse of a right triangle, a straight diagonal from the origin to the point \(\left( {8,7} \right).\) Perhaps you have heard the saying “as the crow flies,” which means the shortest distance between two points because a crow can fly in a straight line even though a person on the ground has to travel a longer distance on existing roadways.
2.1.6 Using the Midpoint Formula
When the endpoints of a line segment are known, we can find the point midway between them. This point is known as the midpoint and the formula is known as the midpoint formula. Given the endpoints of a line segment, \(\left( {x_{1},y_{1}} \right)\) and \(\left( {x_{2},y_{2}} \right),\) the midpoint formula states how to find the coordinates of the midpoint \(M.\)
\[M = \left( {\frac{x_{1} + x_{2}}{2},\frac{y_{1} + y_{2}}{2}} \right)\]
A graphical view of a midpoint is shown in Figure 16. Notice that the line segments on either side of the midpoint are congruent.
Figure 16
Section Exercises
Verbal
1. Is it possible for a point plotted in the Cartesian coordinate system to not lie in one of the four quadrants? Explain.
Solution (click to reveal)
Answers may vary. Yes. It is possible for a point to be on the \(x\)-axis or on the \(y\)-axis and therefore is considered to NOT be in one of the quadrants.
2. Describe the process for finding the \(x\)-intercept and the \(y\)-intercept of a graph algebraically.
3. Describe in your own words what the \(y\)-intercept of a graph is.
Solution (click to reveal)
The \(y\)-intercept is the point where the graph crosses the \(y\)-axis.
4. When using the distance formula \(d = \sqrt{\left( {x_{2} - x_{1}} \right)^{2} + \left( {y_{2} - y_{1}} \right)^{2}},\) explain the correct order of operations that are to be performed to obtain the correct answer.
Algebraic
For each of the following exercises, find the \(x\)-intercept and the \(y\)-intercept without graphing. Write the coordinates of each intercept.
5. \(y = -3x + 6\)
Solution (click to reveal)
The \(x\)-intercept is \(\left( {2,0} \right)\) and the \(y\)-intercept is \(\left( {0,6} \right).\)
6. \(4y = 2x - 1\)
7. \(3x - 2y = 6\)
Solution (click to reveal)
The \(x\)-intercept is \(\left( {2,0} \right)\) and the \(y\)-intercept is \(\left( {0,-3} \right).\)
8. \(4x - 3 = 2y\)
9. \(3x + 8y = 9\)
Solution (click to reveal)
The \(x\)-intercept is \(\left( {3,0} \right)\) and the \(y\)-intercept is \(\left( {0,\frac{9}{8}} \right).\)
10. \(2x - \frac{2}{3} = \frac{3}{4}y + 3\)
For each of the following exercises, solve the equation for \(y\) in terms of \(x\).
11. \(4x + 2y = 8\)
Solution (click to reveal)
\(y = 4 - 2x\)
12. \(3x - 2y = 6\)
13. \(2x = 5 - 3y\)
Solution (click to reveal)
\(y = \frac{5 - 2x}{3}\)
14. \(x - 2y = 7\)
15. \(5y + 4 = 10x\)
Solution (click to reveal)
\(y = 2x - \frac{4}{5}\)
16. \(5x + 2y = 0\)
For each of the following exercises, find the distance between the two points. Simplify your answers, and write the exact answer in simplest radical form for irrational answers.
17. \((-4,1)\) and \((3,-4)\)
Solution (click to reveal)
\(d = \sqrt{74}\)
18. \((2,-5)\) and \((7,4)\)
19. \((5,0)\) and \((5,6)\)
Solution (click to reveal)
\(d = \sqrt{36} = 6\)
20. \((-4,3)\) and \((10,3)\)
21. Find the distance between the two points given using your calculator, and round your answer to the nearest hundredth.
\((19,12)\) and \((41,71)\)
Solution (click to reveal)
\(d \approx 62.97\)
For each of the following exercises, find the coordinates of the midpoint of the line segment that joins the two given points.
22. \((-5,-6)\) and \((4,2)\)
23. \((-1,1)\) and \((7,-4)\)
Solution (click to reveal)
\(\left( {3,\frac{- 3}{2}} \right)\)
24. \((-5,-3)\) and \((-2,-8)\)
25. \((0,7)\) and \((4,-9)\)
Solution (click to reveal)
\(\left( {2,-1} \right)\)
26. \((-43,17)\) and \((23,-34)\)
Graphical
For each of the following exercises, identify the information requested.
27. What are the coordinates of the origin?
Solution (click to reveal)
\(\left( {0,0} \right)\)
28. If a point is located on the \(y\)-axis, what is the \(x\)-coordinate?
29. If a point is located on the \(x\)-axis, what is the \(y\)-coordinate?
Solution (click to reveal)
\(y = 0\)
For each of the following exercises, plot the three points on the given coordinate plane. State whether the three points you plotted appear to be collinear (on the same line).
30. \(\left( {4,1} \right)\left( {-2,-3} \right)\left( {5,0} \right)\)
31. \((-1,2)(0,4)(2,1)\)
Solution (click to reveal)
not collinear
32. \(\left( {-3,0} \right)\left( {-3,4} \right)\left( {-3,-3} \right)\)
33. Name the coordinates of the points graphed.
Solution (click to reveal)
\({\text{A:}\mspace{9mu}\left( {-3,2} \right)},{\text{B:}\mspace{9mu}\left( {1,3} \right)},{\text{C:}\mspace{9mu}\left( {4,0} \right)}\)
34. Name the quadrant in which the following points would be located. If the point is on an axis, name the axis.
ⓐ \((-3,-4)\) ⓑ \((-5,0)\) ⓒ \((1,-4)\) ⓓ \((-2,7)\) ⓔ \((0,-3)\)
For each of the following exercises, construct a table and graph the equation by plotting at least three points.
35. \(y = \frac{1}{3}x + 2\)
Solution (click to reveal)
| \(x\) | \(y\) |
| \(-3\) | 1 |
| 0 | 2 |
| 3 | 3 |
| 6 | 4 |
Table 5
36. \(y = -3x + 1\)
37. \(2y = x + 3\)
Solution (click to reveal)
| \(x\) | \(y\) |
| –3 | 0 |
| 0 | 1.5 |
| 3 | 3 |
Table 6
Numeric
For each of the following exercises, find and plot the \(x\)- and \(y\)-intercepts, and graph the straight line based on those two points.
38. \(4x - 3y = 12\)
39. \(x - 2y = 8\)
Solution (click to reveal)
40. \(y - 5 = 5x\)
41. \(3y = -2x + 6\)
Solution (click to reveal)
42. \(y = \frac{x - 3}{2}\)
For each of the following exercises, use the graph in the figure below.
43. Find the distance between the two endpoints using the distance formula. Round to three decimal places.
Solution (click to reveal)
\(d = 8.246\)
44. Find the coordinates of the midpoint of the line segment connecting the two points.
45. Find the distance that \(\left( {-3,4} \right)\) is from the origin.
Solution (click to reveal)
\(d = 5\)
46. Find the distance that \(\left( {5,2} \right)\) is from the origin. Round to three decimal places.
47. Which point is closer to the origin?
Solution (click to reveal)
\(\left( {-3,4} \right)\)
Technology
For the following exercises, use your graphing calculator to input the linear graphs in the Y= graph menu.
After graphing it, use the 2nd CALC button and 1:value button, hit enter. At the lower part of the screen you will see “x=” and a blinking cursor. You may enter any number for \(x\) and it will display the \(y\) value for any \(x\) value you input. Use this and plug in \(x\) = 0, thus finding the \(y\)-intercept, for each of the following graphs.
48. \(\text{Y}_{1} = -2x + 5\)
49. \(\text{Y}_{1} = \frac{3x - 8}{4}\)
Solution (click to reveal)
\(x = 0\mspace{9mu}\text{~~~~~~~~}y = \operatorname{}-2\)
50. \(\text{Y}_{1} = \frac{x + 5}{2}\)
For the following exercises, use your graphing calculator to input the linear graphs in the Y= graph menu.
After graphing it, use the 2nd CALC button and 2:zero button, hit ENTER. At the lower part of the screen you will see “left bound?” and a blinking cursor on the graph of the line. Move this cursor to the left of the \(x\)-intercept, hit ENTER. Now it says “right bound?” Move the cursor to the right of the \(x\)-intercept, hit ENTER. Now it says “guess?” Move your cursor to the left somewhere in between the left and right bound near the \(x\)-intercept. Hit ENTER. At the bottom of your screen it will display the coordinates of the \(x\)-intercept or the “zero” to the \(y\)-value. Use this to find the \(x\)-intercept.
Note: With linear/straight line functions the zero is not really a “guess,” but it is necessary to enter a “guess” so it will search and find the exact \(x\)-intercept between your right and left boundaries. With other types of functions (more than one \(x\)-intercept), they may be irrational numbers so “guess” is more appropriate to give it the correct limits to find a very close approximation between the left and right boundaries.
51. \(\text{Y}_{1} = -8x + 6\)
Solution (click to reveal)
\(x = 0.75y = 0\)
52. \(\text{Y}_{1} = 4x - 7\)
53. \(\text{Y}_{1} = \frac{3x + 5}{4}\) Round your answer to the nearest thousandth.
Solution (click to reveal)
\(x = - 1.667y = 0\)
Extensions
54. Someone drove 10 mi directly east from their home, made a left turn at an intersection, and then traveled 5 mi north to their place of work. If a road was made directly from the home to the place of work, what would its distance be to the nearest tenth of a mile?
55. If the road was made in the previous exercise, how much shorter would the person’s one-way trip be every day?
Solution (click to reveal)
\(15 - 11.2 = 3.8\text{mi}\) shorter
56. Given these four points: \(A(1,3)\), \(B(–3,5)\), \(C(4,7)\), and \(D(5,–4)\) find the coordinates of the midpoint of line segments \(\overline{\text{AB}}\) and \(\overline{\text{CD}}.\)
57. After finding the two midpoints in the previous exercise, find the distance between the two midpoints to the nearest thousandth.
Solution (click to reveal)
\(\text{6}.0\text{42}\)
58. Given the graph of the rectangle shown and the coordinates of its vertices, prove that the diagonals of the rectangle are of equal length.
59. In the previous exercise, find the coordinates of the midpoint for each diagonal.
Solution (click to reveal)
Midpoint of each diagonal is the same point \((2,2)\). Note this is a characteristic of rectangles, but not other quadrilaterals.
Real-World Applications
60. The coordinates on a map for San Francisco are \((53,17)\) and those for Sacramento are \((128,78)\). Note that coordinates represent miles. Find the distance between the cities to the nearest mile.
61. If San Jose’s coordinates are \((76,–12)\), where the coordinates represent miles, find the distance between San Jose and San Francisco to the nearest mile.
Solution (click to reveal)
37mi
62. A small craft in Lake Ontario sends out a distress signal. The coordinates of the boat in trouble were \((49,64)\) One rescue boat is at the coordinates \((60,82)\) and a second Coast Guard craft is at coordinates \((58,47)\). Assuming both rescue craft travel at the same rate, which one would get to the distressed boat the fastest?
63. A person on the top of a building wants to have a guy wire extend to a point on the ground 20 ft from the building. To the nearest foot, how long will the wire have to be if the building is 50 ft tall?
Solution (click to reveal)
54 ft
64. If we rent a truck and pay a $75/day fee plus $.20 for every mile we travel, write a linear equation that would express the total cost per day \(y,\) using \(x\) to represent the number of miles we travel. Graph this function on your graphing calculator and find the total cost for one day if we travel 70 mi.