We have seen how to obtain the equation of a straight line through two given data points by writing and solving a system of two simultaneous linear equations.

Suppose we only have one point, but know the slope. We can then write y = mx + b, substituting for m the known value of the slope. We can then substitute the x and y coordinates of the known point, leaving only b unknown. We can easily solve the resulting equation for b.

For example, suppose we know that when x = 1, y = 12. Suppose we also have information that tells us that m = .5.

We can then write y = .5 x + b, and substitute (1, 12) for (x, y) to obtain 12 = .5(1) + b.

We easily solve this equation for b to obtain b = 11.5.

As always we substitute m and b back into the form y = mx + b, and our equation becomes y = .5 x + 11.5.

Exercises 1 -3

1. If we know that (3, -4) is a data point on a line with slope 2.8, what is the equation of the line?

2. It is possible to shape a container in such a way that when water flows out through a hole in the bottom of the container, the depth vs. time function is linear. Find the linear function depth(t) = mt + b for such a container if it is known that at clock time 30 seconds the depth is 50 centimeters, and that depth is changing at a rate of -.35 centimeters/second.

3. If a door spring has a length that changes linearly with the weight supported as long as the spring's length is at least 40 cm, what is the linear function weight(load) that models the length, provided that the length is 60 cm when the load is 5 pounds, and that the length changes at a rate of 5 cm / pound?

The situation in which we know that a straight line passes through a given data point with a given slope is depicted below. We have to call the given point something, so we call it (x1, y1). Similarly we call the slope m.

The point (x, y) denotes any point on the graph of the line. This point could be close to (x1, y1), far from (x1, y1), to the right or left of (x1, y1). It had to be drawn somewhere for the sake of the picture, but its location could be anywhere on the straight line.

In the next picture we see that the rise from the known point (x1, y1) to the general point (x, y) is (y - y1) and the run is (x - x1). The slope is therefore (y - y1) / (x - x1).

The slope of course has a known value which we represent here by m. Since the expression (y - y1) / (x - x1) also represents the slope, we have m = (y - y1) / (x - x1).

In the figure below we obtain this slope = slope equation m = (y - y1) / (x - x1). We want the y = mx + b form of this equation. We therefore solve the equation for y:

m = (y - y1) / (x - x1) <the original equation>

m (x - x1) = y - y1 <multiplying both sides by (x - x1) >

mx - m x1 = y - y1 <using the Distributive Law>

y = mx - m x1 + y1 <switching sides and adding y1>

y = mx + b, for b = -m x1+ y1 <letting b = -m x1 + y1>

In the last step we use b to stand for the expression -m x1 + y1, which is just a number whenever m and (x1, y1) are given as numbers.

We have let (x, y) stand for any point on the line, and have seen that the coordinates of this point must satisfy the equation y = mx + b for b = -m x1 + y1.

4.  In the above figure (x,y) lies to the right of (x1, y1) and the slope of the line is positive.  Sketch and label a similar figure, but with (x, y) lying to the left and the line having negative slope.

5. Suppose a line passes through (3, -2) with slope 5. What is the slope = slope equation (y - y1) / (x - x1) = m when we substitute these values? What do we get when we solve for y?

Substitute the above values x = 3, y = -2 and m = 5 into the form y = mx + b. Solve for b. Then substitute m and b into the form y = mx + b to obtain the equation of the line.

Compare the equations you obtained from the different methods.

Make a statement about whether or not the two methods seem to give the same results.

6. For practice use the slope = slope formulation to obtain the equation of lines passing through the given points with the given slopes:

through (-3, 2) with slope -4

through (4, 5) with slope -.003

through (-7, -2) with slope 1350

through (19, -12) with slope .54.

7. Use the slope = slope formulation to find the linear function streamRange(t) for the range of the water stream flowing from the side of a uniform cylinder, if the stream range is 50 centimeters at clock time t = 20 seconds, and if the stream range changes by -10 centimeters over a period of 5 seconds. Use your function to find the clock time at which the stream range first falls to 12 centimeters.

8. Assume that the average class grade in this course is a linear function of how many hours of intelligent out-of-class effort students put into it, on the average. Suppose that an average of 40 hours of such effort was sufficient to make an average grade, on a 4-point scale, of 1.2, and that each additional hour added .045 points to this average. Find the linear function pointAverage(aveHours) that models this situation.

Use this model to determine the expected point average if the average number of hours is 90, which is what is expected for a respectable college class.

Determine the average number of hours that would correspond to a 3.0 grade average.

What factors might make the actual point average function nonlinear?

We have shown that if (x, y) is on the line, y = mx + b for b = -m x1 + y1.

We haven't shown that if y = mx + b for b = -m x1 + y1, the point (x, y) is on the line.

This is similar to the logical situation that arises from the statement 'if it's a fish, it swims'. As far as the author knows this is a perfectly true statement. However if we turn the statement around to say 'if it swims, it's a fish' is not true. In fact this statement would be somewhat offensive to the author, who swims regularly but isn't even cold-blooded, except perhaps when grading tests. Being a fish and a being swimmer are not equivalent states of being.

We want to be able to say that (x, y) being on the line of slope m through (x1, y1) is in fact equivalent to satisfying the equation y = mx + b for the specified value of b. This statement is stronger than either of the preceding two statements about this line, since it combines both statements.

To show that y = mx + b for b = -m x1 + y1, we have to show that whenever (x, y) satisfies the equation, the slope from (x1, y1) to (x, y) must be m. We proceed by the following sequence of statements:

If (x, y) satisfies y = mx + b, then y = mx - m x1 + y1, since b = -m x1 + y1.

This means that (x, y) = (x, mx - m x1 + y1).

The rise from (x1, y1) to (x, mx - m x1 + y1) is y2 - y1 = mx - m x1 + y1 - y1 = mx - m x1.

The run from (x1, y1) to (x, mx - m x1 + y1) is x - x1.

The slope is therefore (y - y1) / (x - x1) = (mx - m x1) / (x - x1).

By the distributive law mx - m x1 = m(x - x1).

The slope is therefore m(x - x1) / (x - x1) = m.

We have started with the assumption that (x, y) satisfies the equation, and have concluded that the slope from (x1, y1) to (x, y) must therefore be m. So (x, y) is on the line through (x1, y1) with slope m.

We have just proved what is called a theorem. This theorem says that

Theorem: The point (x, y) lies on the line through (x1, y1) having slope m if, and only if (x, y) is a solution to the equation y = mx + b, with b = -m x1 + y1.

A theorem is a precise mathematical statement that can be proved from the definitions of the terms in its statement, from certain axioms about how numbers behave when they are multiplied, divided, etc., and the rules of logic.

Theorems are good for several things. For one thing they provide statements that anyone who knows enough mathematics can validate, and are accepted by all mathematicians. So they can be used with confidence. Another attribute of most theorems is that they are useful.

The above theorem is useful because it tells us that the set of all points on a certain line which is defined by its geometry (its slope and location) is identical to the set of all points satisfying a certain equation. There aren't any points satisfying the equation that aren't on the line, and there is no point on the line that doesn't satisfy the equation.

We could ask whether there exists any point (x, y) not on the line, but satisfying the equation y = mx + b for b = -m x1 + y1.

Exercises 9-10 

9. Find the equation of the straight line through the t = 5 sec and the t = 7 sec points of the quadratic function depth(t) = .01 t^2 - 2t + 100, where depth is in centimeters when time is in seconds.

What does the slope of your line tell you about the depth function?

Evaluate both the linear function and the quadratic depth function at t = 3, 4, 5, 6, 7, and 8. How closely does the linear function approximate the quadratic function at each of these times?

Relative to the t = 5 and t = 7 points, and excluding these points, at what t value do we obtain the closest values?

On which side of the t = 5 and t = 7 points is the linear approximation closer to the quadratic function? On which side does the quadratic function 'curve away' from the linear most rapidly?

10. Logic question: If I make the statement 'if you work hard in this course, you will make a respectable grade' to everyone in the class, under which of the following circumstances (there may be more than one) would I be proven a liar?