Algebra Know-It-ALL

634 Worked-Out Solutions to Exercises: Chapters 11 to 19

3. We know the coordinates of at least one point (two, actually) and we also know the
   slope. Let’s use the point (2, 2) as the starting basis. We’ve determined that the slope
   is 8/3. The general PS equation, using u and v as the variable names rather than the
   familiarx and y, is

v−v 0 =m(u−u 0 )

Plugging in 2 for v 0 , 8/3 for m, and 2 for u 0 , we have

v− 2 = (8/3)(u− 2)

That’s the PS form of the equation.

4. To get the equation in SI form, we can manipulate the PS equation we obtained in the
   previous solution. That equation, again, is

v− 2 = (8/3)(u− 2)

Using the distributive law for multiplication over subtraction on the right-hand side, we obtain

v− 2 = (8/3)u− 16/3

Adding 2 to the left side, and adding 6/3 (which is equal to 2) to the right side, we get

v= (8/3)u− 16/3 + 6/3

which simplifies to

v= (8/3)u− 10/3

That’s the SI form of the equation.

5. The simplest possible way to graph this equation is to plot the two points we were
   originally given, and then draw a straight line through them. This is done in Fig. B-6.
   The slope, m, is 8/3 as we derived it. The v-intercept, b, is −10/3 as we derived it.


6. The first equation is in SI form. We can tell, by looking at this equation, that the slope
   of the line will be 1 when we graph it. The second equation can be put into SI form by
   considering the subtraction of s as the addition of its negative, getting

t= 5 + (−s)

We can apply the commutative law on the right side to get

t=−s+ 5

Now we can see that the slope of this line will be −1 when we graph it.

In a Cartesian plane where both axes are graduated in increments of the same size, a

slope of 1 corresponds to a ramp angle of 45°, and a slope of −1 corresponds to a ramp