Now let’s use the PS form to derive an equation for the line. We have two points to choose from. Either point
will work. Let’s use (x 1 ,y 1 ). We can substitute x 1 for x 0 , and y 1 for y 0 in the classical PS equation to get
y − y 1 =m(x − x 1 )
Substituting (y 2 − y 1 )/(x 2 − x 1 ) for m, we obtain
y − y 1 = [(y 2 − y 1 )/(x 2 − x 1 )] (x − x 1 )
which can be rearranged to
y − y 1 = (x − x 1 )(y 2 − y 1 )/(x 2 − x 1 )
This is a mess, but it’s the best we can do when we aren’t given the slope directly. The good news is that
most of the values in this equation are constants. When the coordinates for the points are given as numbers,
we can plug them in and get the equation by means of straightforward arithmetic.
Practice Exercises
This is an open-book quiz. You may (and should) refer to the text as you solve these problems.
Don’t hurry! You’ll find worked-out answers in App. B. The solutions in the appendix may not
represent the only way a problem can be figured out. If you think you can solve a particular
problem in a quicker or better way than you see there, by all means try it!
+x
+y
y
x
m= y x
(x,y 11 )
(x,y 22 )
x = + x
y = + y
x
y
1
1
2
2
Figure 15-8 A two-point form of a linear equation can be
derived from this generic graph.
Practice Exercises 249
