Lines & Slope

Slope: what it actually means

A line in the plane is a set of points. What distinguishes it from other curves? A defining property: the rate of change of $y$ with respect to $x$ is constant. Pick any two points on the line, compute $\frac{\Delta y}{\Delta x}$, and you get the same number. That number is the slope, $m$.

$m = \frac{y_2 - y_1}{x_2 - x_1}, \quad x_1 \ne x_2.$

This is "rise over run": the change in height divided by the change in horizontal position.

Why is it the same between any two points? Because we said it is — that's what defines a line. Any set of points with constant $\frac{\Delta y}{\Delta x}$ is a line. Try the slope-explorer widget below to see this live: no matter which two points you drag, the slope triangle always reports the same ratio.

Drag the points around. Note that the line equation updates in real time. The slope never changes as long as you stay on the line — that constancy IS the line.

Slope-intercept form: $y = mx + b$

If a line has slope $m$ and crosses the $y$-axis at $(0, b)$, then for any point $(x, y)$ on the line:

$\frac{y - b}{x - 0} = m \implies y - b = mx \implies y = mx + b.$

This is slope-intercept form. $m$ is the slope; $b$ is the $y$-intercept (the $y$-value when $x = 0$). Reading $y = 3x - 2$: slope 3 (rise 3, run 1), crosses $y$-axis at $(0, -2)$.

Point-slope form: $y - y_1 = m(x - x_1)$

What if you know the slope $m$ and a point $(x_1, y_1)$ but not the intercept? For any other point $(x, y)$ on the line, slope constancy demands:

$\frac{y - y_1}{x - x_1} = m \implies y - y_1 = m(x - x_1).$

This is point-slope form. You can always convert to slope-intercept by solving for $y$.

Horizontal and vertical lines

Horizontal line through $(0, k)$: $y = k$ for every $x$. Slope $= \frac{0}{\Delta x} = 0$. The line goes nowhere vertically.

Vertical line through $(h, 0)$: $x = h$ for every $y$. Slope would be $\frac{\Delta y}{0}$ — division by zero, undefined. Vertical lines have NO slope (not zero slope — undefined slope). They cannot be written as $y = mx + b$; their equation is simply $x = h$.

Parallel and perpendicular: the slope relationships

Parallel lines have equal slopes and different $y$-intercepts. They never intersect because their rates of change are identical.

Perpendicular lines have slopes that satisfy $m_1 \cdot m_2 = -1$, i.e., $m_2 = -\frac{1}{m_1}$. Why? Here's the picture-proof in words: if line 1 has slope $m_1 = \frac{a}{b}$ (rise $a$, run $b$), rotate that slope triangle 90° clockwise. The "rise" becomes $-b$ and the "run" becomes $a$, giving slope $\frac{-b}{a} = -\frac{1}{m_1}$. The product is $\frac{a}{b} \cdot \frac{-b}{a} = -1$.

Example: if one line has slope $\frac{2}{3}$, a perpendicular line has slope $-\frac{3}{2}$.

The big foreshadow

The word "linear" shows up in linear equations, linear inequalities, linear functions, linear systems, linear transformations, and the entire subject of linear algebra. Every single one of those concepts is describing, in some form, the constant-rate-of-change structure you see here. A line is the simplest geometric object with that structure. When we say a function is "linear," we mean it behaves like a line. When we say a transformation is "linear," we mean it respects the same underlying structure. File this away. It will pay dividends for the rest of the course.