Equations of lines

If you have worked through the notes on slope, distance and midpoint, you already know a lot about the algebra of lines. Now we can wrap it up and figure out how to represent a line on a plane with an equation.

Points on a line

Only two properties are needed to fully describe any line in a plane.

A line in the x-y plane (or any plane) is fully described if we know how much it is tilted from horizontal or vertical, and where it is located.

We specfiy tilt by measuring slope, which we defined as the change in the y coordinate divided by the change in x:

$$m = \frac{y_2 - y_1}{x_2 - x_1}$$

We could easily define slope as $\Delta x / \Delta y$, but the way we do it is a long-standing convention. A positive slope means the line runs from lower left to upper right; a negative slope means the opposite.

All that remains is to specify where the line is on the plane. That turns out to be pretty easy; as we move the line up and down (±y), it simultaneously moves left and right (±x). That means we only need to have the location of a single point to specify the location.

Again by long-standing convention, we usually pick the place where the line intersects the y-axis, the y-intercept, where x = 0.

So we generally find the equation that represents a line by finding its slope and its y-intercept. The two mainly-used forms of any equation that models a line are shown below. The slope is as defined above and the y-intercept, $b$, is the point $(0, b)$.

Correspondences

It's not too difficult at all to find correspondences between these two forms of the equation of a line. After all, they represent the same thing. Take the standard form, for example:

$$Ax + By = C$$

We can begin to morph this equation into slope-intercept form by moving the x-term to the right by subtraction:

$$By = -Ax + C$$

Then if we divide both sides by $A$, we get

$$y = \frac{-A}{B} \, x + \frac{C}{B}$$

Now it's pretty easy to see the correspondences between the standard and slope-intercept forms.

Linear transformations

We can think of lines in a different way, as functions of the independent variable, $x$.

$$y = mx + b$$

where the slope, $m$, is a vertical scaling parameter, which stretches or shrinks the grap of the line vertically, and $b$ is a vertical translational parameter. It moves the graph up and down. In the end, that's all we need to fully describe a line in a plane: Where is it located — that's the y-intercept, just a point that pins the line to some location on the pland, and how steeply it's sloped.

You can move the sliders under the graph to change those parameters between -10 and 10 to see what effect they have on the graph.

Notice that for a line, translating it up and down is equivalent to moving it right or left. You can see that very clearly if you just twiddle the y-intercept slider; the line will also appear to move right ↔ left, depending on how you look at it.

In function notation, a linear function looks like

$$f(x) = mx + b$$

and the function $f(x) = x$ is the parent function of all lines, the simplest line we can draw.

Example 1

Find the equation of a line from two points

Here are our two points. They could be any two points on the x-y plane; only one line can be drawn through any two points.

$$(-3, -1) \; \text{ and } \; (2, 7)$$

There are two ways to find the equation. Different people have different opinions about which is easier. In the first, start by finding the slope:

$$m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{7 - (-1)}{2 - (-3)} = \frac{8}{5}$$

Now we need to find the y-intercept, $(b)$. To do that, rearrange the slope-intercept formula by subtracting $mx$ from both sides:

$$y = mx + b \; \longrightarrow \; b = y - mx$$

Now we use our slope and just one of the points. Either will work, but I'm going to choose one with positive components. That's always a good choice – less possibility of silly errors.

So the intercept is (note the I've converted 7 to 5^ths to add the fractions):

$$b = \frac{35}{5} - \frac{16}{5} = \frac{19}{5}$$

Then the formula, in slope-intercept form, is

$$y = \frac{8}{5} x + \frac{19}{5}$$

That's fine, but it's possible to manipulate it so that it has no fractions, just by multiplying both sides by 5:

$$5y = 8x + 19$$

Example 2

Find the equation of a line from one point and the slope

Let's say that our line passes through the point (-2, 3) and has a slope of m = ⁷/₃. This problem is really just like the one in example 1, except that we can skip the step of calculating the slope because we already know it.

We again have two ways to find the equation of this line. The first is just to use the slope-intercept formula to find the y-intercept:

$$y = mx + b \; \longrightarrow \; b = y - mx$$

We use the coordinates of the known point (-2, 3) and the slope to find the y-intercept:

Watch the negative signs! If we convert 3 to 3^rds, we get

$$b = \frac{9}{3} + \frac{14}{3} = \frac{23}{3}$$

Now we just plug the slope and intercept into the slope-intercept formula, y = mx + b:

$$y = \frac{7}{3} + \frac{23}{3}$$

Example 3

Find the equation of a line parallel to another

Problem: Find the equation of a line passing through (1, 2) and parallel to the line passing through points (-1, -1) and (2, 5).

Our first task is to find the slope of the line for which we have two points:

$$ \begin{align} m = \frac{y_2 - y_1}{x_2 - x_1} &= \frac{5 - (-1)}{2 - (-1)} \\ \\ &= \frac{6}{3} = 2 \end{align}$$

Because parallel lines have the same slope, the slope of our new line is m = 2. We can plug that slope and our point (1, 2) into the point-slope formula:

That gives:

$$y - 2 = 2(x - 1)$$

A little rearrangement gives us

$$y = 2x - 2 + 2$$

... and finally a very simple equation for our function:

$$y = 2x$$

Example 4

Find the equation of a line perpendicular to another

Problem: Find the equation of a line that passes through the point (-1, 4) and is perpendicular to the line passing through points (2, 7) and (-1, -3).

Here again (see example 3) we first calculate the slope of the line for which we have enough information:

$$ \begin{align} m = \frac{y_2 - y_1}{x_2 - x_1} &= \frac{-3 -7}{-1 -2} \\ \\ &= \frac{-10}{-3} = \frac{10}{3} \end{align}$$

Now the slope of a line perpendicular to that line is

$$m_{\perp} = \frac{-3}{10}$$

Now we can plug that slope and the one point we know that lies along our new line, (-1, 4), into the point-slope formula:

That give us

$$y - 4 = \frac{-3}{10} (x + 1)$$

multiplying through and moving the 4 to the right by addition (note that I've converted it to 40/10 so I have a common denominator), we get

$$y = \frac{-3}{10} x - \frac{3}{10} + \frac{40}{10}$$

Practice problems

A. Find the equation of the line passing through these pairs of points:

1.   (-2, -2) and (6, 4)

2.   (-3, 7) and (4, 2)

3.   (1, 5) and (5, 1)

B. Find an equation for the lines passing through the point given and with the specified slope, m.

7.   (-3, 6), m = 5/9

8.   (22, 4), m = -7/26

A quick word on lines in more than one plane ...

Take a look at the two lines in the figure to the left. The x-y plane is meant to be in the plane of your screen, and the y-z plane is sticking out of it (there's also an x-z plane I haven't shown).

The black line is in the y-z plane, and the magenta line is in the x-y plane. They do not cross, and never do, yet they are not parallel.

We have to be careful about our understanding of geometry problems.