Straight Lines

Concept	Formula
Slope	$m={\displaystyle \frac{y_2-y_1}{x_2-x_1}}={\displaystyle \frac{\text{rise}}{\text{run}}}$
Point-slope form	$y-y_1=m(x-x_1)$
Parallel lines	$m_1=m_2$
Perpendicular lines	$m_1{m}_2=-1$

Slope of a Line

Consider a straight line $L$ and two distinct points $P(x_1,y_1)$ and $Q(x_2,y_2)$ on it. The slope of the line, denoted by $m$, is defined to be the ratio:

$$m=\frac{y_2-y_1}{x_2-x_1}=\frac{\text{rise}}{\text{run}}.$$

If $\alpha$ () is the angle that a non-horizontal line makes with the positive direction of the $x$-axis, then $m=\tan \alpha$. (For a quick review of trigonometric functions, see Section 1.17.)

The sign of the slope is related to the direction of the line:

• If , $\alpha$ is an acute angle and the line rises to the right.
• If , $\alpha$ is an obtuse angle and the line falls to the right.
• If $m=0$, the line is horizontal.

Equations of a Line

The equation of the line passing through a given point $P(x_1,y_1)$ with slope $m$ is:

$$y-y_1=m(x-x_1).$$

This is called the point-slope form of the equation of the line.

If we are given two points $(x_1,y_1)$ and $(x_2,y_2)$, we first calculate $m=(y_2-y_1)/(x_2-x_1)$, then substitute into the point-slope form to obtain the point-point form:

$$y=y_1+\left(\frac{y_2-y_1}{x_2-x_1}\right)(x-x_1).$$

• The equation of a vertical line through $(x_1,y_1)$ is simply $x=x_1$. A vertical line has no defined slope (the run is zero, so the rise-over-run ratio is undefined).

Parallel and Perpendicular Lines

Two lines with slopes $m_1$ and $m_2$ are:

• parallel $⟺$ $m_1=m_2$.
• perpendicular $⟺$ $m_1{m}_2=-1$.

Horizontal lines (slope 0) are perpendicular to vertical lines (undefined slope), which is consistent with the above rule in the limiting sense.

Frequently Asked Questions

What does the slope of a tangent line mean in calculus?

The derivative f'(a) is defined as the slope of the tangent line to the graph of $y=f(x)$ at the point $(a,f(a))$. It measures the instantaneous rate of change of $f$ at $x=a$. The equation of the tangent line at that point is y - f(a) = f'(a)(x-a), which is just the point-slope form.

How do I find the equation of a line perpendicular to $y=2x+1$ passing through $(3, 0)$?

The slope of $y=2x+1$ is $m_1=2$. A perpendicular line has slope $m_2=-\frac{1}{m_1}=-\frac{1}{2}$. Using the point-slope form with point $(3,0)$: $$y-0=-\frac{1}{2}(x-3)⟹y=-\frac{1}{2}x+\frac{3}{2}.$$

What if a line has slope zero, is it still a function?

Yes. A horizontal line $y=c$ is a perfectly valid function (a constant function). A vertical line $x=k$, however, is not a function, because it assigns multiple $y$-values to the single $x$-value $k$.