SLOPE1 program for the TI-83

Note that the TI-83 and the TI-83 plus have slightly different keys. This page uses the keys associated with the TI-83. The differences are that the TI-83 key is replaced by the TI-83 Plus key, and the TI-83 key is replaced by the TI-83 Plus key.

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 SLOPE1 program on the TI-83 implements such an algorithm. Note that the SLOPE1 program uses the TOSTR program in order to generate nice looking output. The TI-83 version of the SLOPE1 program is available as slope1.83p. The version of TOSTR, presented in the 208304.htm page, is available as tostr.83p.

Figure 1 We can start by pressing the key to open the PRGM menu. Then we press the key to move the highlight selector down to the SLOPE1 program. The calculator used to generate Figure 1 holds a large number of programs. Figure 1 shows the screen at the point where we have identified the SLOPE1 program.
Figure 2 Once the SLOPE1 program was identified in Figure 1, we press the key to select that program and to paste prgmSLOPE1 as a command on the screen, as shown in Figure 2.
We can execute the command and start running the program by pressing the key.

Figure 3 In Figure 3, the program has started. 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 give X2 the value 10, and to leave the screen as it appears in Figure 3.

Figure 4 We moved from Figure 3 to Figure 4 by pressing the key, and then waiting for the calculator to do its work. In response the calculator computes the slope, given as ^–1 in Figure 4, and it then continues to generate the slope-intercept form of the resulting equation.
At that point the program is in a "paused" condition, waiting for us to press the ENTER key to continue. The "moving" line of dots in the upper right corner of the screen indicate that "paused" condition.
We press to move to Figure 5.

Figure 5 The program continues by computing and displaying the STANDARD form of the resulting equation. Again, the program is "paused" at this point.
We press to move to Figure 6.

Figure 6 Figure 6 again shows the program in a "paused" condition, where the program has indicated that it is done with the current problem and it is waiting to continue with the next problem.
We press to move to Figure 7.

Figure 7 The program has started over, from the beginning. The screen has been cleared, the heading has been rewritten, and the program prompts for values of the coordinates of our two points. 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 8.

Figure 8 The calculator has determined the slope and the slope-intercept form of the solution. From the "paused" condition of Figure 8 we press the key to move to Figure 9.

Figure 9 Here the calculator has determined the standard form of the solution equation.
Please note that the SLOPE1 program produces a standard form version of the equation that does not require the leading coefficient to be non-negative.
Press the key to move to Figure 10.

Figure 10 The calculator has completed the problem, and it is ready to move on to the next problem. Press the key to move to Figure 11.

Figure 11 Figure 11 represents the data input for yet another example: "Find the equation of the line containing the points (2,7) and (5,9)."

Figure 12 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 13 Figure 13 contains the standard form of the solution equation.

Figure 14 The calculator has completed the problem, and it is ready to move on to the next problem. Press the key to move to Figure 15.

Figure 15 Figure 15 represents the data input for yet another example: "Find the equation of the line containing the points (^–4,3) and (^–7,25)."

Figure 16 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 17 Figure 17 contains the standard form of the solution equation.

Figure 18 The calculator has completed the problem, and it is ready to move on to the next problem. Press the key to move to Figure 19.

Figure 19 Figure 19 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. Let us see how the SLOPE1 program responds to these values.

Figure 20 The SLOPE1 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 21 Figure 21 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. Let us see how the SLOPE1 program responds to these values.

Figure 22 The SLOPE1 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 23 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 23. There are only two options. Either we can GOTO or QUIT. We want to QUIT, so we press the key to move to Figure 24.

Figure 24 Figure 24 looks remarkably similar to Figure 22. However, in Figure 22 the calculator was in a "paused" condition. In Figure 24 we are out of the program and ready to do anything that we want to do.

The SLOPE1 program is designed to generate an equation if we are given the coordinates of two points on the line. How can we use the SLOPE1 program to solve problems such as

Find the equation of the line that has slope=4/7 and that contains the point (9,6).

In order to use the SLOPE1 program, we need to know two points on the line. The problem statement gives us one point, (9,6). Can we get a second point? Yes. We know the slope and we know that

M =	4	=	change in y

	7		change in x

Therefore, if we start at the point (9,6) and we add 7 to the X value and add 4 to the Y value we produce a new point (16,10) 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 SLOPE1 program.

Figure 25 Figure 25 shows that the program has been restarted. Having broken out of the program in Figures 23 and 24, we only needed to press the key to re-issue the the last command, namely prgmSLOPE1, and therefore restart the program. In Figure 25 we have entered the two points derived from the discussion above. We can press to accept the value for y₂ and to move to Figure 26.

Figure 26 The first thing to notice in Figure 26 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 form for the desired equation.

Figure 27 Figure 27 gives the standard form of the equation.

Figure 28 The calculator has completed the problem, and it is ready to move on to the next problem. Press the key to move to Figure 29.

Figure 29 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 SLOPE1 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 SLOPE1 program. The values for the two points have been entered into the program in Figure 29.

Figure 30 Figure 30 confirms the second point by calculating the expected slope. Then, the SLOPE1 program produces the slope-intercept form of the desired equation.

Figure 31 Figure 31 gives the standard form of the equation.

Figure 32 The calculator has completed the problem, and it is ready to move on to the next problem. Press the key to move to Figure 33.

Figure 33 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 SLOPE1 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 SLOPE1 program. The values for the two points have been entered into the program in Figure 33.

Figure 34 Figure 34 confirms the second point by calculating the expected slope. Then, the SLOPE1 program produces the slope-intercept forms of the desired equation.

Figure 35 Figure 35 gives the standard form of the equation.

Figure 36 The calculator has completed the problem, and it is ready to move on to the next problem.

Figure 1	We can start by pressing the key to open the PRGM menu. Then we press the key to move the highlight selector down to the SLOPE1 program. The calculator used to generate Figure 1 holds a large number of programs. Figure 1 shows the screen at the point where we have identified the SLOPE1 program.
Figure 2	Once the SLOPE1 program was identified in Figure 1, we press the key to select that program and to paste prgmSLOPE1 as a command on the screen, as shown in Figure 2. We can execute the command and start running the program by pressing the key.
Figure 3	In Figure 3, the program has started. 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 give X2 the value 10, and to leave the screen as it appears in Figure 3.
Figure 4	We moved from Figure 3 to Figure 4 by pressing the key, and then waiting for the calculator to do its work. In response the calculator computes the slope, given as ^–1 in Figure 4, and it then continues to generate the slope-intercept form of the resulting equation. At that point the program is in a "paused" condition, waiting for us to press the ENTER key to continue. The "moving" line of dots in the upper right corner of the screen indicate that "paused" condition. We press to move to Figure 5.
Figure 5	The program continues by computing and displaying the STANDARD form of the resulting equation. Again, the program is "paused" at this point. We press to move to Figure 6.
Figure 6	Figure 6 again shows the program in a "paused" condition, where the program has indicated that it is done with the current problem and it is waiting to continue with the next problem. We press to move to Figure 7.
Figure 7	The program has started over, from the beginning. The screen has been cleared, the heading has been rewritten, and the program prompts for values of the coordinates of our two points. 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 8.
Figure 8	The calculator has determined the slope and the slope-intercept form of the solution. From the "paused" condition of Figure 8 we press the key to move to Figure 9.
Figure 9	Here the calculator has determined the standard form of the solution equation. Please note that the SLOPE1 program produces a standard form version of the equation that does not require the leading coefficient to be non-negative. Press the key to move to Figure 10.
Figure 10	The calculator has completed the problem, and it is ready to move on to the next problem. Press the key to move to Figure 11.
Figure 11	Figure 11 represents the data input for yet another example: "Find the equation of the line containing the points (2,7) and (5,9)."
Figure 12	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 13	Figure 13 contains the standard form of the solution equation.
Figure 14	The calculator has completed the problem, and it is ready to move on to the next problem. Press the key to move to Figure 15.
Figure 15	Figure 15 represents the data input for yet another example: "Find the equation of the line containing the points (^–4,3) and (^–7,25)."
Figure 16	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 17	Figure 17 contains the standard form of the solution equation.
Figure 18	The calculator has completed the problem, and it is ready to move on to the next problem. Press the key to move to Figure 19.
Figure 19	Figure 19 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. Let us see how the SLOPE1 program responds to these values.
Figure 20	The SLOPE1 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 21	Figure 21 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. Let us see how the SLOPE1 program responds to these values.
Figure 22	The SLOPE1 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 23	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 23. There are only two options. Either we can GOTO or QUIT. We want to QUIT, so we press the key to move to Figure 24.
Figure 24	Figure 24 looks remarkably similar to Figure 22. However, in Figure 22 the calculator was in a "paused" condition. In Figure 24 we are out of the program and ready to do anything that we want to do.