Lines

Section 7.2 Lines

Subsection 7.2.1 Coordinates and Equations

Figure 7.2.1. Lines

Section 7.1 reviewed or introduced the Cartesian plane. Once we have coordiantes, we can use them to describe geometric objects. The first object we want to describe are straight lines.

By convention, the first horizontal coordinate is called the \(x\)-coordinate and the second horizontal coordinate is called the \(y\)-coordinate. If we have an algebraic equation in the variables \(x\) and \(y\text{,}\) then we can interpret that equation as a geometric shape. Let me explain with an example.

\begin{equation*} y = \frac{1}{2} x + 2 \end{equation*}

This is an equation in the two variables. For any point \((x,y)\text{,}\) we can ask whether the point satisfies the equation by substituting it in and seeing if the right side and left side match. Try the point \((4,0)\) by replacing the \(x\) with \(4\) and the \(y\) with \(0\text{.}\)

\begin{equation*} 0 = \frac{1}{2} 4 + 2 \end{equation*}

The left side is \(0\) and the right side works out to \(4\text{.}\) Since \(0\) is not equation to \(4\text{,}\) the point \((4,0)\) does not satisfy the equation, therefore will not be on the corresponding shape. Now try the point \((-2,1)\text{.}\)

\begin{equation*} 1 = \frac{1}{2} (-2) + 2 \end{equation*}

The left side is \(1\) and the right side works out to \(-1 + 2 = 1\text{.}\) Both give the same value, so this point satisfies the equation. Therefore \((-2,1)\) will be a point on the corresponding shape.

It turns out there are infinitely many points that will satisfy this equation and that, if we draw all these points, we get a line. Figure 7.2.2 shows the line with several points indicated. You are free to check that all the label points do, in fact, satisfy the equation of the line.

Figure 7.2.2. The Line \(y = \frac{1}{2} x + 2\)

Subsection 7.2.2 Equations of Lines

Figure 7.2.3. Equations of Lines

We can make many equations in the variales \(x\) and \(y\text{,}\) giving many types of shapes in the plane. In this review, we are interested only in straight lines. The equation we used an exanmple above gave a line, but we would like to know how to describe any line.

Any equation where the variables \(x\) and \(y\) show up simply multiplied by a constant and then added or subtracted from another constant produces a line. By simply, I mean that no other operations (other than multiplication by a constant) are performed on the variables: no exponents, no roots, no functions, nothing else. These equations, since they produce lines, are called linear equations. Here are some examples.

\begin{align*} y \amp = 7x - 9 \\ 5y + 4x - 10 \amp = 0\\ \frac{1}{5} y - \frac{2}{7} x - \frac{5}{17} \amp = 0 \\ \sqrt{7}y \amp = \pi x + \frac{1}{\sqrt{19}} \end{align*}

In all these equations, all we have done is multiplied each varaible by come constant. Then we can add and subtract the variables, possibly with another constant, but no other operationrs are performed.

If we wanted a general form, then let the symbols \(a_1\text{,}\) \(a_2\) and\(a_3\) stand for constants. We could bring everything over to one side of the equation and write the general form of an equation of a line this way:

\begin{equation*} a_1 x + a_2 y + a_3 = 0 \end{equation*}

This is a reasonable general form. All lines in the plane can be described by these equations. However, this is not the most intuitive form. I'm going to work towards another form, which may be familiar to you. To get there, we need to review another important concept.

Subsection 7.2.3 Slope

The slope of a straight line is a measure of its steepness. A large slope descrbes a steep line -- one that is quickly increasing. A small slope describes a shallow line -- one that is slowly increasing. A negative slope describes a decreasing line.

A line with a slope of zero is a horizontal line: it does not increase or decrease at all. A vertical line has no slope at all: we say that its slope is undefined.

To calculate the slope, we take choose any two points on the line. On the line from Section 7.1, we could choose the point \((2,3)\) and \((6,5)\text{.}\) We calculate the difference in the \(y\) coordinates and call this the rise: it measures how far up the line goes between the two points. Then we calculate the difference in the \(x\) coordinates and call this the run: it measures how far we move horizontally between the two points. The slope is then the rise divided by the run.

\begin{equation*} \text{Slope} = \frac{\text{Rise}}{\text{Run}} = \frac{5-3}{6-2} = \frac{2}{4} = \frac{1}{2} \end{equation*}

This line has slope \(\frac{1}{2}\text{.}\) The geometry of this calculation is show in Figure 7.2.4

Figure 7.2.4. Slope of the Line \(y = \frac{1}{2} x + 2\)

This method of calculation shows us again why veritcal lines have no slope. The \(x\) coordinates of any two points on a vertical line are the same, to the run would be zero. To calculate the slope would involve division by zero, which is not defined. Therefore, vertical lines have no slope.

Subsection 7.2.4 Slope Intercept Form

Now that we have defined slope, we can review/introduce a special way of writing the equations of lines. Go back to the four example lines above; in each of these, I will re-arrange the equation so that \(y\) is along on the left side.

\begin{align*} y \amp = 7x - 9 \\ y \amp = - \frac{4}{5} x + 2\\ y \amp = \frac{10}{7} x + \frac{25}{17} \\ y \amp = \frac{\pi}{\sqrt{7}} x + \frac{1}{\sqrt{19}\sqrt{7}} \end{align*}

I can describe the general form we get by solving for \(y\) on the left. Let \(m\) and \(b\) represent constant (the choice of these letters if conventional and used in many places). Then all of these equations have this form:

\begin{equation*} y = mx + b \end{equation*}

In this form, the number \(m\) will always be the slope of the line. The number \(b\) will be the \(y\)-coordinate of the point where the line crosses the \(y\) axis; this is called the \(y\)-intercept. These two pieces of information are enough to complete determine the line: if we know where it crosses the \(y\)-axis and we know how quickly is grows from that point (its slope), then we know everything we need to determine the line. Figure 7.2.5 shows an example to illustrate this slope-intercept form.

Figure 7.2.5. Slope-Intercept Form

Before finishing this section, I want to briefly talk about two special cases. First, a horizontal line has a slope of zero. Therefore, in slope-intercept form, \(m = 0\text{,}\) so the equation of the line look like

\begin{equation*} y = 0x + b = b \end{equation*}

Horizontal lines have equations \(y = b\) for some constant. All we need to specify if the \(y\) coordinate. \(y = 4\) is the horizontal line where the \(y\)-coordinate is always \(4\text{,}\) and the \(x\)-coordinate can be anything whatsoever. Figure 7.2.6 shows some horizontal lines.

Figure 7.2.6. Horizontal Lines

We've said several times that a vertical line has no slope and cannot be expressed in slope-intercept form. However, the previous paragraph gives us a good idea of how to understand vertical lines. If a horizontal line is simply determined by its \(y\) coordinate, a vertical line is similarly determined by its \(x\) coordinate. Any line of the form \(x = b\) is a horizontal line. \(x - -3\) is the horizontal line of points with \(x\)-coordinate \(-3\) and any \(y\) coordinate whatsoever. Figure 7.2.7 shows some horizontal lines.

Figure 7.2.7. Vertical Lines

Subsection 7.2.5 Calcluating Equations of Lines

We often need to calculate the equation of a line gives various pieces of information about this line.

If we are give the slope and the \(y\)-intercept, we just put those numbers directly into the slope-intercept form. For example, if we wanted the line with slope \(\frac{-2}{5}\) and \(y\)-intercept \(3\text{,}\) we would simply put those numbers into the form and get the equation of the line
\begin{equation*} y = \frac{-2}{5} x + 3 \end{equation*}
Sometimes we are given a point and a slope. For example, say we wanted the equation of the line with slope \(4\) that goes through the point \((2,2)\text{.}\) We have the slope, so we can put that in place of \(m\) in the slope intercept form.
\begin{equation*} y = 4x + b \end{equation*}
The \(y\)-intercept is still unknown. However, if we put in the point, we can solve for \(b\text{.}\)
\begin{align*} y \amp = 4x + b \\ 2 \amp = (4)(2) + b \\ 2 \amp = 8 + b \\ -6 \amp = b \end{align*}
That gives us a value for the \(y\)-intercept, hence the equation of the line.
\begin{equation*} y = 4x - 6 \end{equation*}
This process works for any line give a slope and a point: put the slope into the slope-intercept form, put the point in for the coordinates \(x\) and \(y\text{,}\) and solve for the intercept.
There is one last way of describing a line: by giving two point. There is a unique line passing between any two points -- but how do we get the equation of this line? We calculate the slope. Recall that slope is defined as the fraction rise over run. Rise is the different in the \(y\) coordinates and run is the different in the \(x\) coordinates, so give two points we can calculate the slope. I'll show this by example: what is the line through the point \((3,-5)\) and \((1,1)\text{?}\) First I'll calculate the slope
\begin{equation*} m = \frac{\text{Rise}}{\text{Run}} = \frac{-5 - 1}{3 - 1} = \frac{-6}{2} = -3 \end{equation*}
The slope is \(-3\text{,}\) so the slope intercept looks like
\begin{equation*} y = -3x + b \text{.} \end{equation*}
Then the process is the same as the previous case: we have the slope and we have a point (two, in fact). We put in a point (either point will work) to calculate the intercept.
\begin{align*} y \amp = -3x + b \\ -5 \amp = (-3)(3) + b \\ -5 \amp = -9 + b \\ 4 \amp = b \end{align*}
With the intercept, we can write the equation of the line.
\begin{equation*} y = -3x + 4 \end{equation*}
I'll leave it up to you to check that if I had used the other point, I would have calculated the same value for \(b\text{.}\)