The slope program for the TI-89 (92)

We are often given the problem of finding the equation of a line given two distinct points on the line. The solution is almost a mechanical process. The following flowchart illustrates the steps:

Such an algorithm can be programmed into the calculator. The slope program on the TI-89 (or on the TI-92) implements such an algorithm. The TI-89 version of the slope program is available as slope.89p and the TI-92 version of the slope program is available as slope.92p.

Figure 1	We can start by pressing the keys to open the VAR-LINK menu. The calculator used to generate Figure 1 holds a large number of programs. Within folders the program and variable names are displayed in alphabetic order. We are looking for the "slope" program, and we can have the display shift to the next item that starts with the letter "s" by pressing the key. In fact, that key was pressed twice to arrive at the image shown in Figure 1.
Figure 2	Figure 2 reflects the condition of having found the slope program in the VAR-LINK window, and of pressing the key to select that program and to paste its name into the command entry and edit line of the HOME screen. Actually, what gets pasted there is "slope(". We need to complete the command, to enter a right parenthesis, by pressing the key. We can execute the command and start running the program by pressing the key.
Figure 3	In Figure 3, the program has started. Output from the program appears in the PROGRAM IO window. The program clears the screen and displays "Get Slope, etc." on the top line, followed by a prompt for the value of "x1" on the second line. We are looking for the solution to "Find the equation of the line containing the points (4,7) and (10,1)." Therefore, we respond with for x1. The program accepts that value and continues by asking for the value for y1. assigns 7 to y1. This is followed by to leave the screen as it appears in Figure 3. At this point we are in danger of losing the top portion of the screen if we go any further.
Figure 4	We press to give x2 the value 10. Then we press to create the value for y2. To accpet that value and move to Figure 5 we press the key.
Figure 5	The slope program responds by repeating the input points, computing and displaying the slope of the line (in this case, – 1), computing and displaying both the slope-intercept form and the standard form of the desired equation. Having produced all of the answers, the program is in a "paused" condition, as indicated by the displayed in the lower right corner of the screen. We can leave this screen and continue with the next set of input points by pressing the key.
Figure 6	Rather than present screen images that show the data input, we will skip to the main output screen. In this case we are solving the problem: "Find the equation of the line containing the points (5,1) and (8,10)." Once the values have been given, we press to accept the final value and move to Figure 6. The calculator has determined the slope, the slope-intercept form of the solution, and the standard form of the solution. Please note that the slope program produces a standard form version of the equation that does not require the leading coefficient to be non-negative.
Figure 7	Figure 7 follows from the data input for yet another example: "Find the equation of the line containing the points (2,7) and (5,9)." The calculator has determined the slope between the two points to be 2/3. Note that the calculator program has taken the liberty of writng the slope-intercept form as y=2/3 x + 17/3. It would have been better to have included parentheses so that the answer appears as y=(2/3) x + 17/3, leaving no doubt but that the "x" is not in the denominator of the fraction. Once we see and understand the somewhat "sloppy" form that the program uses for fractional slopes, we can correctly understand the equations that the program produces.
Figure 8	Figure 8 follows from the data input for yet another example: "Find the equation of the line containing the points (– 4,3) and (– 7,25)." The calculator has determined the slope between the two points to be – 22/3. Note that in the slope-intercept form of the solution equation, the calculator program has left the slope and the second coordinate of the y-intercept as improper fractions. slope does not convert these to either mixed numbers or to decimal values.
Figure 9	Figure 9 represents the data input for yet another example: "Find the equation of the line containing the points (2,9) and (8,9)." By inspection we can see that the solution will be a horizontal line. The slope program correctly identifies the equation as the constant function, with the solution equation being y=9. Note that we could write this as y=0x+9 to produce the strict slope intercept form. In addition we could write this as 0x + 1y = 9 to produce the strict standard form of the equation.
Figure 10	Figure 13 represents the data input for yet another example: "Find the equation of the line containing the points (6,1) and (6,10)." By inspection we can see that the solution will be a vertical line. The slope program correctly identifies the equation as a vertical line with the solution equation being x=6. Note that we could write this as 1x + 0y = 6 to produce the strict standard form of the equation. It is impossible to construct a slope-intecept form of this equation: the slope is undefined and the coefficient of y is 0.
Figure 11	If we want to stop the program we need to break out of it. We can press the key to break out of the program. The result is shown in Figure 11. There are only two options. Either we can GOTO or QUIT. We want to QUIT, so we press the key to move to Figure 12.
Figure 12	Figure 12 looks remarkably similar to Figure 10. However, in Figure 10 the calculator was in a "paused" condition. In Figure 12 the program has terminated. However, we are still in the Program IO screen. We will need to press the key to move back to the HOME screen, shown in Figure 13.
Figure 13	Figure 13 shows the updated HOME screen. Note tht we have a record of having run the slope program and of having terminated that program through a "break". In addition, the previous command, slope(), remains in the command input and edit line.

The SLOPE program is designed to generate an equation if we are given the coordinates of two points on the line. How can we use the SLOPE program to solve problems such as

Figure 14	Figure 14 shows that the program has been restarted. Having broken out of the program in Figures 11 through 13, we only needed to press the key to re-issue the the last command, namely slope(), and therefore restart the program. In Figure 14 we have entered the two points derived from the discussion above. The first thing to notice in Figure 14 is that the slope has been computed to be exactly the value given in the problem statement. This is verification that we have correctly computed the coordinates of the second point. Then we can look at the slope-intercept and standard forms for the desired equation.
Figure 15	The next problem to consider is Find the equation of the line that has slope=2 and that contains the point (– 4,1). In order to use the slope program, we need to know two points on the line. The problem statement gives us one point, (– 4,1). Can we get a second point? Yes. We know the slope and we know that m =     2     =     change in y     1     change in x   Therefore, if we start at the point (– 4,1) and we add 1 to the x value and add 2 to the y value we produce a new point (– 3,3) that must be on the desired line. Thus, we use the definition of slope to produce a second point, and this allows us to use the slope program. The values for the two points have been entered into the program to arrive at Figure 15. The screen confirms the second point by calculating the expected slope. Then, the slope program produces the slope-intercept and the standard forms of the desired equations.
Figure 16	The next problem to consider is Find the equation of the line that has slope=– 3/5 and that contains the point (– 6,– 13). In order to use the slope program, we need to know two points on the line. The problem statement gives us one point, (– 6,– 13). Can we get a second point? Yes. We know the slope and we know that m =     – 3     =     change in y     5     change in x   Therefore, if we start at the point (– 6,– 13) and we add 5 to the x value and add – 3 to the y value we produce a new point (– 1,– 16) that must be on the desired line. Thus, we use the definition of slope to produce a second point, and this allows us to use the slope program. The values for the two points have been entered into the program in Figure 16. In fact, Figure 16 shows the state of the screen after we have entered the value for y2, but before we have accpeted that value. We can observe the coordinates of the second point in Figure 16. We press the key to accept that value and move to Figure 17.
Figure 17	Figure 20 confirms the second point by calculating the expected slope. Then, the SLOPE program produces the slope-intercept and the standard forms of the desired equations. Notice that the coordinates of the second point have been cut off on the right side of the screen.