# 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 $y = \displaystyle \frac{4}{5}x - \frac{6}{5}$. My 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 $(\overline{x}, \overline{y})$.

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