Lesson 4-5: Slope-Intercept Form

Property

The $x$-intercept is the point $(a, 0)$ where the line crosses the $x$-axis.

The $y$-intercept is the point $(0, b)$ where the line crosses the $y$-axis.

The $x$-intercept occurs when $y$ is zero.

Property

For a line L, for any two points P, Q on the line, the quotient is constant, and that constant is the slope of the line.

This creates a test for any point $(x, y)$ to be on the line. If we know the slope $m$ and one point $(x_1, y_1)$ on the line, any other point $(x, y)$ must satisfy the equation:

Multiplying both sides by $(x - x_1)$ gives the point-slope form $y - y_1 = m(x - x_1)$, which simplifies to the slope-intercept form $y = mx + b$.

Examples

• A line has a slope of $4$ and passes through the point $(1, 7)$. Any other point $(x, y)$ on the line must satisfy $\frac{y-7}{x-1} = 4$. Multiplying by $(x-1)$ gives $y-7 = 4(x-1)$, which simplifies to the equation $y = 4x + 3$.
• To find the equation of a line passing through $(2, 5)$ and $(4, 11)$, first find the slope: $m = \frac{11-5}{4-2} = \frac{6}{2} = 3$. Using the point $(2, 5)$, the equation is $\frac{y-5}{x-2} = 3$, which simplifies to $y = 3x - 1$.
• Let's check if the point $(5, 14)$ is on the line $y = 3x - 1$. We substitute $x=5$ into the equation: $y = 3(5) - 1 = 15 - 1 = 14$. Since the calculated y-value matches the point's y-value, the point is on the line.

Explanation

Think of a line's equation as its membership rule. If you know the slope (the line's steepness) and one point on it, any other point can only 'join' the line if it maintains that exact same steepness with the known point.