Slope Form and Two-Point Equation

The most useful forms for writing the equation of a line are those that make the geometric meaning of the parameters visible — the slope tells you how steeply the line rises, and specifying two points pins down the line uniquely.

Slope and Angle of Inclination

📐Definition — Slope of a Line

The slope (or gradient) of a non-vertical line $\ell$ is the number

$k = \operatorname{tg}\alpha$

where $\alpha$ is the angle of inclination — the angle measured counterclockwise from the positive $x$ -axis to the line, with $0° \leq \alpha < 180°$ and $\alpha \neq 90°$ .

Equivalently, for any two distinct points $(x_1, y_1)$ and $(x_2, y_2)$ on $\ell$ with $x_1 \neq x_2$ :

$k = \frac{y_2 - y_1}{x_2 - x_1} \quad \text{("rise over run")}$

Sign of the slope:

$k > 0$ : the line rises from left to right ( $0° < \alpha < 90°$ )
$k < 0$ : the line falls from left to right ( $90° < \alpha < 180°$ )
$k = 0$ : the line is horizontal ( $\alpha = 0°$ )
Vertical lines have no slope (undefined, $\alpha = 90°$ )

Slope-Intercept Form

⚡Theorem — Slope-Intercept Form (Theorem 12.1)

Every non-vertical line with slope $k$ and $y$ -intercept $b$ (the point where it crosses the $y$ -axis) has equation:

$\boxed{y = kx + b}$

Conversely, every equation of this form represents a non-vertical line with slope $k$ and $y$ -intercept $b$ .

Proof. A line with slope $k$ passing through $(0, b)$ : for any point $(x, y)$ on it (with $x \neq 0$ ), the slope condition gives $\frac{y - b}{x - 0} = k$ , so $y - b = kx$ , i.e., $y = kx + b$ . When $x = 0$ : $y = b$ $\checkmark$ . $\blacktriangleleft$

Point-Slope Form

If a line has slope $k$ and passes through a known point $(x_0, y_0)$ , then any other point $(x, y)$ on the line satisfies:

$\boxed{y - y_0 = k(x - x_0)}$

This follows immediately from the definition of slope. Expanding gives $y = kx + (y_0 - kx_0)$ , i.e., slope-intercept form with $b = y_0 - kx_0$ .

Two-Point Equation

⚡Theorem — Two-Point Equation (Theorem 12.2)

The line through two distinct points $A(x_1, y_1)$ and $B(x_2, y_2)$ with $x_1 \neq x_2$ has equation:

$\boxed{\frac{y - y_1}{y_2 - y_1} = \frac{x - x_1}{x_2 - x_1}}$

Proof. A point $P(x, y)$ lies on line $AB$ if and only if $\overrightarrow{AP}$ and $\overrightarrow{AB}$ are collinear (parallel), i.e., their slopes are equal:

$\frac{y - y_1}{x - x_1} = \frac{y_2 - y_1}{x_2 - x_1}$

Cross-multiplying gives the two-point equation. $\blacktriangleleft$

Note: If $x_1 = x_2$ (vertical line), the equation is simply $x = x_1$ .

Parallel and Perpendicular Lines

⚡Theorem — Parallelism and Perpendicularity (Theorem 12.3)

Let $\ell_1$ and $\ell_2$ be two non-vertical lines with slopes $k_1$ and $k_2$ . Then:

$\ell_1 \parallel \ell_2$ (parallel, distinct) $\iff k_1 = k_2$ (and the lines have different $y$ -intercepts)
$\ell_1 \perp \ell_2$ (perpendicular) $\iff k_1 \cdot k_2 = -1$ , equivalently $k_2 = -\dfrac{1}{k_1}$

Proof of perpendicularity. If $\ell_1$ has inclination angle $\alpha_1$ , then $\ell_2 \perp \ell_1$ means $\alpha_2 = \alpha_1 + 90°$ (or $\alpha_1 - 90°$ ). Using the tangent of a sum:

$k_2 = \operatorname{tg}(\alpha_1 + 90°) = -\cot\alpha_1 = -\frac{1}{\operatorname{tg}\alpha_1} = -\frac{1}{k_1} \qquad \blacktriangleleft$

Worked Examples

✎Example — Example 1 — Line Through Two Points

Write the equation of the line through $A(-1, 3)$ and $B(3, -1)$ .

Solution. Compute the slope:

$k = \frac{-1 - 3}{3 - (-1)} = \frac{-4}{4} = -1$

Using point-slope form with point $A(-1, 3)$ :

$y - 3 = -1 \cdot (x - (-1)) = -(x + 1)$

$y = -x - 1 + 3 = -x + 2$

Verification: $A(-1, 3)$ : $-(-1) + 2 = 3$ $\checkmark$ . $B(3, -1)$ : $-3 + 2 = -1$ $\checkmark$ .

Answer: $y = -x + 2$ . $\blacktriangleleft$

✎Example — Example 2 — Perpendicular Line

Find the equation of the line through $P(2, 5)$ perpendicular to the line $y = 3x - 4$ .

Solution. The given line has slope $k_1 = 3$ . The perpendicular slope is:

$k_2 = -\frac{1}{k_1} = -\frac{1}{3}$

Using point-slope form with $P(2, 5)$ :

$y - 5 = -\frac{1}{3}(x - 2)$

$y = -\frac{x}{3} + \frac{2}{3} + 5 = -\frac{x}{3} + \frac{17}{3}$

Or equivalently: $x + 3y - 17 = 0$ .

Answer: $y = -\dfrac{x}{3} + \dfrac{17}{3}$ (or $x + 3y - 17 = 0$ ). $\blacktriangleleft$

✎Example — Example 3 — Angle of Inclination

A line passes through the origin and makes an angle of $135°$ with the positive $x$ -axis. Write its equation.

Solution. The slope is $k = \operatorname{tg} 135° = \operatorname{tg}(180° - 45°) = -\operatorname{tg} 45° = -1$ .

The line passes through the origin, so $b = 0$ . Equation: $y = -x$ .

Answer: $y = -x$ . $\blacktriangleleft$

Exercises

✏Exercise

Write the equation of the line:

With slope $k = 2$ passing through $(0, -3)$
Through the points $A(0, 4)$ and $B(6, 0)$
Through $C(-2, 1)$ and $D(4, 4)$

✏Exercise

For each pair of lines, determine whether they are parallel, perpendicular, or neither.

$y = 2x + 1$ and $y = 2x - 5$
$y = 3x$ and $y = -\frac{1}{3}x + 7$
$y = 4x - 2$ and $y = -4x + 2$

Hint: Check slopes — same slope means parallel; product of slopes $= -1$ means perpendicular.

✏Exercise

A triangle has vertices $A(0, 0)$ , $B(4, 0)$ , $C(2, 6)$ .

Find the equation of each side of the triangle.
Find the equation of the median from $C$ to side $AB$ .
Find the equation of the altitude from $C$ to side $AB$ .