Class 4 Notes

I. Finding Equations of Lines
   A. Finding the Equation from a Point and Slope
   B. Point-Slope Form
   C. Finding the Equation from Two Points
   D. Practice Problems
II. Relations Between Lines
   A. Parallel Lines
   B. Perpendicular Lines
   C. Practice Problems


I. Finding Equations of Lines

Last week we learned how to plot lines, how to go from an equation to
a graph.

As we discussed, though, the graph and the equation really represent
the same thing -- so it must be possible to go from graphical
information back to an equation.

That's the first thing we're going to talk about today: how to find
the equation of a line given information about it.


A. Finding the Equation from a Point and Slope

Recall last week we said that if we know the slope and intercept of a
line, we know everything about it.

More generally, if we know the slope and any point on the line, we
know everything about it.

It's easy to see this intuitively: by telling you a point, I tell you
where the line starts.

By telling you the slope, I tell you which direction it goes.

Thus, if I give you a point and a slope, I've told you where to start
and which way to go, which is all there is to a line.

So all we need to do is take the intuitive information and turn it
into a method for finding the equation.

Let's consider an example:

(on board) Find the equation of a line passing through the point (-1,2)
with slope -1.

We'll begin by trying to find the equation in slope-intercept form,
which we saw last time is the most convenient form in general.

(on board) y = mx + b

Recall tht this equation describes a generic line, with m and b as the
slope and y intercept.

We need to "fill in" m and b in this equation so that it describes our
line.

The value of m is easy enough to fill in: we know that the slope is
-1, and m is just the slope. Thus:

(on board) m = -1

We can get b almost as easily.

We know that the point (-1,2) is on the line, meaning that if we put
in -1 for x and 2 for y, the equation has to come out true.

We can just plug in (-1,2) into our y = mx+b form, and that will tell
us what b is.

(on board)
y = mx + b
y = -x + b
2 = -(-1) + b
2 = 1 + b
b = 1

So we just plug in the point we've been given for x and y, and then
solve for b.

Thus, the equation for this line is

y = -x + 1.

To be sure, let's plot this:

x     Equation	       y
0     y = -(0) + 1     1
1     y = -(1) + 1     0
-1    y = -(-1) + 1    2
2     y = -(2) + 1     -1
-2    y = -(-2) + 1    3

(draw graph)

As we can see from looking at the graph or the table, the line does
indeed pass through the point (-1,2).

The slope is downhill, as it should be for a negative slope, and it
has the right steepness. So this works.

Let's try another example.

(on board)
Find the line through the point (2,3) with slope 1/2.

We start with the equation

(on board) y = mx + b

What do we do first? (let class answer)

We immediately know that m = 1/2.

What next? (let class answer)

Then we plug in x and y to get b:

(on board)
y = 1/2 x + b
3 = 1/2 (2) + b
3 = 1 + b
b = 2

So the equation is

(on board)
y = 1/2 x + 2

So the approach we have is very simple:

(on board)
1. Write down y = mx + b.
2. The slope is m.
3. Plug in the point for x and y, and solve for b.


B. Point-Slope Form (10)

There is a slight variation on this method which your book uses to
find the equation of a line from a point and a slope.

I personally think it's fairly silly, since the method we just used is
so simple, but let's talk about it for a moment anyway.

Consider the following way of writing the equation of a line:

(on board)
y - y1 = m(x - x1)

where x1, y1, and m are just some numbers.

The first thing to notice here is that this is the equation of a
line.

We could easily re-arrange it into the form something times x plus
something times y equal some number.

The key ingredients are there: a number times x, a number times y, and
other pure numbers.

Why would we choose to write the equation of a line in this somewhat
bizarre form?

Here's why: if we're given a point and a slope, we can plug them
directly into this form.

Let's consider our last problem:

(on board)
Find the equation of the line that passes through (2,3) with slope
1/2.

Now let's write down the point-slope formula:

(on board)
y - y1 = m (x - x1)

Let's fill in the slope for m, and put in (2,3) for x1 and y1.

Then we get:

(on board)
y - 3 = 1/2 (x - 2)

How can we simplify this into y = mx + b form? (let class answer)

(on board)
y - 3 = 1/2 x - 1
y = 1/2 x + 2

Notice that's the same equation we got before.

So this is another method of finding the equation given a point and a
slope:

(on board)
1. Write down y - y1 = m (x - x1).
2. The slope is m.
3. Plug in the point for x1 and y1.
4. Simplify.


C. Finding the Equation from Two Points

We can extend this a step further.

Suppose we have two points.

(draw two points on the board)

It's pretty clear that there's only one straight line we can draw
between these two points.

(draw a line through the points)

So there must be a way to go from the coordinates of two points to the
equation of the line joining them.

In fact, the way is really simple.

Recall from last week that we know how to find the slope between two
points.

To find the equation of the line through two points, then, all we have
to do is find the slope between them and use our method for finding
the equation given a point and a slope.

Let's do an example:

(on board)
Find the equation of the line through (2,3) and (-2,1).

OK, first we find the slope. Who remembers how to do that? (let class
answer)

(on board)
m = delta y / delta x
  = (y2 - y1) / (x2 - x1)
  = (1 - 3) / (-2 - 2)
  = -2 / (-4)
  = 1/2

Now we just find the equation using the slope and one of the points --
it doesn't matter which one, as they'll both give the same answer.

How do we do this? (let class answer)

(on board)
y = 1/2 x + b
3 = 1/2 (2) + b
3 = 1 + b
b = 2

So the equation of the line is

(on board)
y = 1/2 x + 2

Just to verify, let's try plugging in the other point:

(on board)
y = 1/2 x + b
1 = 1/2 (-2) + b
1 = -1 + b
b = 2

So we get the same equation:

(on board)
y = 1/2 x + 2

Let's graph this.

(on board)
x     Equation		 y
0     y = 1/2 (0) + 2	 2
1     y = 1/2 (1) + 2	 5/2
-1    y = 1/2 (-1) + 2	 3/2
2     y = 1/2 (2) + 2	 3
-2    y = 1/2 (-1) + 2	 1

(draw graph)

As you can see, this line does indeed pass through both the given
points.

So finding the line through two points is very simple: just find the
slope, then find the line using the point and slope with either
point.


D. Practice Problems

Practice problems to do as time permits:

1. Plot the points (-3,2) and (-1,1). Find the equation of the line
connecting the points, and plot it.

2. An architect is drawing blueprints for a wheelchair ramp. In his
blueprints, he chooses a coordinate system so the base of the ramp is
at (0,0) and y is the vertical direction.
a. The wheelchair ramp is to have a slope of 1/10. Find the equation
of the line describing the ramp.
b. Plot the line.
c. The ramp must climb a height of 1 foot. Find the coordinates of the
top of the ramp.


II. Relations Between Lines

A. Parallel Lines

Something else neat we can do is use the equations to look at
relationships between lines.

One common occurence is parallel lines, lines that run next to each
other, at constant distance, never touching or diverging.

The idea of parallelism gives us another way of defining a line.

Suppose I give you a line and a point not on that line.

(draw on board)

As you can see intuitively, there is only one way to draw a line
through the point so that it is parallel to the first line.

(draw on board)

Thus, this situation uniquely defines a line.

There must therefore be a way of getting from a line and point to find
the equation of the parallel line -- and there is.

Let's take a concrete example.

(on board)
Find the line through (1,2) parallel to the line y = x - 1.

The key thing to notice is that parallel lines have to be equally
steep -- otherwise they'd converge or diverge.

Thus, parallel lines must have the same slope!

With this insight, it's easy to see how to proceed.

What is the slope of y = x - 1? (let class answer)

It's just 1.

Thus, the line we're looking for must also have slope 1.

We now have a point and a slope, so we've reduced this to a problem we
know how to solve: finding the equation given a point and a slope.

How do we proceed then? (let class answer)

We know that m is 1, so we just have to plug in x and y to find b.

(on board)
y = mx + b
y = x + b
2 = 1 + b
b = 1

So the equation of the line is

(on board)
y = x + 1

Notice that the parallel lines has the same "y = mx" part of its
equation, but that the "b" is different. That is always true of
parallel lines.

Also note that we can do this in reverse, and use this as a check as
to whether two lines are parallel.

Consider:

(on board)
Is the line connecting (-3,-1) to (1,2) parallel to the line
connecting (-2,0) to (2,2)?

To check this, we simply compute the slopes of the two lines.

How do we do this? (let class answer)

(on board)
m1 = (y2 - y1) / (x2 - x1)
   = (2 - (-1)) / (1 - (-3))
   = 3 / 4

m2 = (y2 - y1) / (x2 - x1)
   = (2 - 0) / (2 - (-2))
   = 1 / 2

The slopes are different, to the lines are not parallel. If they had
come out the same, that would mean the lines were parallel.


B. Perpendicular lines

If we can find parallel lines, we can also find perpendicular lines.

Again, consider a line and a point.

(draw picture)

Suppose I tell you I want a line that passes through the point and is
perpendicular to the first line.

(draw picture)

As you can see, that again uniquely defines a line. Thus, there must
be a way to find the equation of that line.

Let's take another concrete example:

(on board)
Find the line through (1,1) perpendicular to y = 2x - 1.

To see how to approach this problem, let's graph the line we've been
given.

(on board)
x	Equation	y
0	y = 2(0) - 1	-1
1	y = 2(1) - 1	1
-1	y = 2(-1) - 1	-3

(draw graph)

We can envision what the perpendicular line will have to look like.

(draw picture)

First of all, notice that, since the original line is uphill, the
perpendicular line will have to be downhill. If the original line had
been downhill, the perpendicular line would be uphill.

If we say this in terms of slopes, the slope of the perpendicular line
has to have the opposite sign of the slope of the original line.

Similarly, if the original line is steep, the perpendicular line will
have to be a gradual slope, and vice versa.

So if the original line has a big positive slope, for example, its
perpendicular will have to have a small (in the absolute value sense)
negative slope.

It turns out that the rule is this:

(on board)
If a line has slope m, its perpendicular has slope m_perp = -1/m.

In other words, the slopes of perpendicular lines are negative
reciprocals of one another.

So if m is positive, the perpendicular has a negative slope. And if m
is big, the perpendicular will be small, which is exactly what we
want.

Let's apply this to our example.

What is the slope of the original line? (let class answer)

It's just 2.

So the slope of the perpendicular is

m_perp = -1/2.

Now we're back to the problem we know how to solve: finding the
equation given the slope and a point: (1,1).

How do we find the equation? (let class answer)

We know the slope is -1/2, so we write down:

(on board)
y = (-1/2) x + b
1 = (-1/2)(1) + b
1 = -1/2 + b
b = 3/2

So the perpendicular equation is

(on board)
y = (-1/2) x + 3/2

Let's graph this just to be sure:

x     Equation		    y
0     y = (-1/2)(0) + 3/2   3/2
1     y = (-1/2)(1) + 3/2   1
-1    y = (-1/2)(-1) + 3/2  2

(draw graph)
As we can see, this is just like what we think it should look like.

As with parallel lines, we can turn this knowledge around and use it
as a test to see if two lines are perpendicular.

Consider this problem:

(on board)
Are 4x - 2y = 3 and x = 1 - 2y perpendicular?

How should we approach this? (let class answer)

We just find the slope of each line. It's easiest to do that by
putting them into slope-intercept form.

(on board)
4x - 2y = 3
-2y = -4x + 3
y = 2x - 3/2

So this line has slope 2.

(on board)
x = 1 - 2y
1 - 2y = x
-2y = x - 1
y = -1/2 x + 1/2

So this line has slope -1/2.

Clearly, 2 times -1/2 is -1, so the lines are perpendicular.


C. Practice Problems

As time permits:

1a. Plot y = (1/3) x + 1
 b. Find the equation of the line parallel to this through (0,-1), and
    plot the parallel line.
 c. Find the equation of the line perpendicular to the two lines you
    just plotted, passing through (0,-1).

2. Consider the wheelchair ramp from earlier. The next floor up will
   have a ramp in the same place on that floor, 15 feet higher than
   the first ramp.
   a. What is its equation?
   b. Graph the lines describing both ramps.