Slope-intercept form of a line

The slope-intercept form of a line is based on the size of the angle that the line makes with the x-axis.

Motivation

We have a line p with the general equation ax + by + c = 0. We know that the graph of a linear function is a line. We usually write a linear function in the form y = f(x), so we might want to convert the general equation ax + by + c = 0 to this form. To do that we need b ≠ 0, because we have to isolate y, which means dividing the whole equation by b. If b ≠ 0, we can rearrange the general equation like this:

$$\begin{array}{rcl}ax+by+c& =& 0(÷b)\\ \frac{ax}{b}+y+\frac{c}{b}& =& 0(\text{isolate }y)\\ y& =& -\frac{ax}{b}-\frac{c}{b}\end{array}$$

This gives us the usual form of a linear function. Because this expression is rather clumsy, we introduce new letters: we write m for the fraction $-\frac{a}{b}$, which is the slope (gradient) of the line, and b for the fraction $-\frac{c}{b}$. Careful: this new b (the y-intercept) is a different number from the coefficient b of the general equation. Then we can write the previous equation as:

$$y=mx+b$$

It is not a different form: instead of the ugly fraction $-\frac{c}{b}$ we have simply written b, and that is all. Because mx + b describes an ordinary linear function, a line in this form cannot be parallel to the y-axis. We have already ruled that out by requiring the coefficient b of the general equation to satisfy b ≠ 0.

If this coefficient were b = 0, the line would be parallel to the y-axis and the general equation of such a line would have the form ax + c = 0. If we divide this equation by a and rearrange it, we get:

$$\begin{array}{rcl}ax+c& =& 0(÷a)\\ x+\frac{c}{a}& =& 0\\ x& =& -\frac{c}{a}\end{array}$$

Because the fraction $-\frac{c}{a}$ is ugly again, we introduce the substitution $d=-\frac{c}{a}$, so in the end we get the equation

$$x=d$$

To sum up, we can write every line that is not parallel to the y-axis in the form y = mx + b, where m, b ∈ ℝ, and every line parallel to the y-axis in the form x = d, where d ∈ ℝ.

An equation of the form y = mx + b is called the slope-intercept form of a line, and we call the coefficient m the slope of the line.

Example

We have a line that passes through the points A(2, 8), B(8, 10). Find the slope-intercept form of the line. The slope-intercept form is best found from the general equation of the line, so we construct that first. The general equation has the form ax + by + c = 0, where (a, b) is a normal vector, so we find a normal vector. The direction vector has coordinates (8, 10) − (2, 8) = (6, 2), so the normal vector $\overrightarrow{\mathbf{n}}$ has coordinates (−2, 6). We can substitute a = −2, b = 6 into the general equation:

$$-2x+6y+c=0$$

In place of (x, y) we substitute a point that the line certainly passes through, for example the point B(8, 10). So we substitute x = 8, y = 10 into the equation:

$$\begin{array}{rcl}-2\cdot 8+6\cdot 10+c& =& 0\\ -16+60+c& =& 0\\ c& =& -44\end{array}$$

We get the general equation

$$-2x+6y-44=0$$

We can divide this equation by −2 to get a slightly nicer form:

$$x-3y+22=0$$

We get the slope-intercept form by isolating y. We move all the other terms to the right-hand side and then divide the equation by three:

$$\begin{array}{rcl}x-3y+22& =& 0\\ -3y& =& -x-22(\cdot (-1))\\ 3y& =& x+22(\cdot \frac{1}{3})\\ y& =& \frac{1}{3}x+\frac{22}{3}\end{array}$$

The last line is the line in slope-intercept form. If we want the coefficients m, b of the slope-intercept form, we write $m=\frac{1}{3},b=\frac{22}{3}$. You can also look at the graph of the line with the direction vector $\overrightarrow{\mathbf{u}}$ and the normal vector $\overrightarrow{\mathbf{n}}$:

If the coefficient b had not come out as awkward as $\frac{22}{3}$, you would see at a glance that b is the y-coordinate of the point where the line crosses the y-axis (the y-intercept). The point where the line meets the y-axis has x-coordinate x = 0. If we substitute this into the equation $y=\frac{1}{3}x+\frac{22}{3}$, we get $y=\frac{1}{3}\cdot 0+\frac{22}{3}=\frac{22}{3}$, so for x = 0 the value of y is $y=\frac{22}{3}$.

The slope of a line

If we have the equation of a line in slope-intercept form y = mx + b, we call the number m the slope of the line. The slope is related to the angle that the line makes with the positive x-axis. Look at the following picture:

The picture shows four different lines with different slopes m but always the same coefficient b = 2. All the lines have the form y = mx + 2. Notice that as the slope, that is, the number in front of the unknown x, changes continuously, the angle that the line makes with the positive x-axis changes too. For example, the purple line $y=\frac{1}{2}x+2$ makes the angle δ with the axis, the red line y = x + 2 makes the angle γ and the green line y = 2x + 2 makes the angle β.

In this case, the greater the slope, the greater the angle the line makes with the axis. The slope of the purple line $y=\frac{1}{2}x+2$ is $m_1=\frac{1}{2}$, the slope of the green line y = 2x + 2 is m₂ = 2, so m₁ < m₂. At the same time δ < β. But the slope of the blue line y = −2x + 2 is m₃ = −2, which is smaller than all the previous ones, and yet the angle α is the largest of all the angles.

We can see that the slope m is related to the angle somehow. Let’s look at it more closely. We will show only the purple line $y=\frac{1}{2}x+2$, and we will be interested in the angle δ. We choose any two different points A, B on the line and use them to build the right-angled triangle ABC, as shown in the picture. The angles δ and ϵ are equal, because they are corresponding angles.

We will now try to express the angle ϵ using the line segments AC and BC. To do that, we need to know their lengths. Let’s label the coordinates of the points A and B like this: A = (x₁, y₁) = (2, 3) and B = (x₂, y₂) = (6, 5). The length of the segment AC is then equal to the difference |x₂ − x₁| = |6 − 2| = 4. We take the x-coordinate of the point B, which is the same as the x-coordinate of the point C, and subtract the x-coordinate of the point A. This gives us the length of the segment AC. We do the same with the segment BC, only with the y-coordinates. So |BC| = |y₂ − y₁| = |5 − 3| = 2.

Next we use the trigonometric functions, specifically the tangent. In the right-angled triangle ABC, for the angle ϵ we have $\tan (ϵ)=\frac{|BC|}{|AC|}$. The tangent of the angle ϵ is the ratio of the opposite side to the adjacent side. If we substitute the distances for the segments, we get:

$$\tan (ϵ)=\frac{|BC|}{|AC|}=\frac{|y_2-y_1|}{|x_2-x_1|}=\frac{2}{4}=\frac{1}{2}$$

So the tangent of the angle ϵ, which is the same as the angle δ, comes out as $\frac{1}{2}$. The tangent of the angle that the line makes with the positive x-axis is $\frac{1}{2}$. And the equation of this line is $y=\frac{1}{2}x+2$, so the tangent of the angle that the line makes with the positive x-axis is equal to the slope of the line! This is no coincidence, and it holds in general:

If a line p has the slope-intercept equation y = mx + b, then the slope m is equal to the tangent of the angle that the line p makes with the positive x-axis.