Finding the equation of a line between two points

Here’s a standard problem that could be found in any Algebra I textbook.

Find the equation of the line between (-1,-2) and (4,2).

The first step is clear: the slope of the line is

m = \displaystyle \frac{2-(-2)}{4-(-1)} = \frac{4}{5}

At this point, there are two reasonable approaches for finding the equation of the line.

Method #1. This is the method that was hammered into my head when I took Algebra I. We use the point-slope form of the line:

y - y_1 = m (x - x_1)

y - 2 = \displaystyle \frac{4}{5} (x-4)

y - 2 = \displaystyle \frac{4}{5}x - \frac{16}{5}

y = \displaystyle \frac{4}{5}x - \frac{6}{5}

For what it’s worth, the point-slope form of the line relies on the fact that the slope between (x,y) and (x_1,y_1) is also equal to m.

Method #2. I can honestly say that I never saw this second method until I became a college professor and I saw it on my students’ homework. In fact, I was so taken aback that I almost marked the solution incorrect until I took a minute to think through the logic of my students’ solution. Let’s set up the slope-intercept form of a line:

y= \displaystyle \frac{4}{5}x + b

Then we plug in one of the points for x and y to solve for b.

2 = \displaystyle \frac{4}{5}(4) + b

\displaystyle -\frac{6}{5} = b

Therefore, the line is y = \displaystyle \frac{4}{5}x - \frac{6}{5}.

green lineMy experience is that most college students prefer Method #2, and I can’t say that I blame them. The slope-intercept form of a line is far easier to use than the point-slope form, and it’s one less formula to memorize.

Still, I’d like to point out that there are instances in courses above Algebra I that the point-slope form is really helpful, and so the point-slope form should continue to be taught in Algebra I so that students are prepared for these applications later in life.

Topic #1. In calculus, if f is differentiable, then the tangent line to the curve y=f(x) at the point (a,f(a)) has slope f'(a). Therefore, the equation of the tangent line (or the linearization) has the form

y = f(a) + f'(a) \cdot (x-a)

This linearization is immediately obtained from the point-slope form of a line. It also can be obtained using Method #2 above, so it takes a little bit of extra work.

This linearization is used to derive Newton’s method for approximating the roots of functions, and it is a precursor to Taylor series.

Topic #2. In statistics, a common topic is finding the least-squares fit to a set of points (x_1,y_1), (x_2,y_2), \dots, (x_n,y_n). The solution is called the regression line, which has the form

y - \overline{y} = r \displaystyle \frac{s_y}{s_x} (x - \overline{x})

In this equation,

  • \overline{x} and \overline{y} are the means of the x- and y-values, respectively.
  • s_x and s_y are the sample standard deviations of the x- and y-values, respectively.
  • r is the correlation coefficient between the x- and y-values.

The formula of the regression line is decidedly easier to write in point-slope form than in slope-intercept form. Also, the point-slope form makes the interpretation of the regression line clear: it must pass through the point of averages (\overline{x}, \overline{y}).

Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

Connecting to %s

This site uses Akismet to reduce spam. Learn how your comment data is processed.