Write equations of linear functions in slope-intercept and point-slope form given a graph, a slope and a point, or two points, and apply the slope relationships for parallel and perpendicular lines (A.F.4).

What this topic is asking

A.F.4 asks you to write the equation of a linear function from a graph, a slope and a point, or two points, and to apply the parallel and perpendicular slope relationships. On the Virginia Algebra I SOL these are Functions items: write a line in slope-intercept or point-slope form, or find a parallel or perpendicular slope. They appear as fill-in-the-blank, multiple choice, and coordinate-plane items.

The three forms of a line

The Algebra I formula sheet gives three forms, each useful for a different job:

• Slope-intercept $y = mx + b$: best for reading the slope and $y$-intercept, and for graphing.
• Point-slope $y - y_1 = m(x - x_1)$: best for building a line from a slope and a point.
• Standard $Ax + By = C$: a tidy integer form, useful for intercepts.

Writing from a slope and a point

If the point is the $y$-intercept (on the $y$-axis, $x = 0$), read $b$ directly. Otherwise, use point-slope or substitute the point into $y = mx + b$ and solve for $b$.

Writing from two points

Find the slope first with $m = \frac{y_2 - y_1}{x_2 - x_1}$, then use either point in point-slope form (or solve for $b$). For $(1, 2)$ and $(3, 8)$: $m = \frac{8 - 2}{3 - 1} = 3$, then $y - 2 = 3(x - 1)$, giving $y = 3x - 1$.

Parallel and perpendicular slopes

The slope encodes a line's direction, so the relationships between lines are relationships between slopes:

• Parallel lines never meet, so they have the same slope (and different $y$-intercepts). A line parallel to $y = \frac{2}{3}x + 1$ has slope $\frac{2}{3}$.
• Perpendicular lines meet at a right angle, and their slopes are opposite reciprocals: flip the fraction and change the sign. A line perpendicular to slope $\frac{2}{3}$ has slope $-\frac{3}{2}$. The product of perpendicular slopes is $-1$.

Why perpendicular slopes are opposite reciprocals

The opposite-reciprocal rule is not arbitrary, it comes from rotating a line by a right angle. A slope of $\frac{a}{b}$ means "go $b$ across and $a$ up." Rotating that direction $90$ degrees turns "across" into "up" and "up" into "across," and one of them reverses direction, so the new step is "go $a$ across and $b$ down" (or up), a slope of $-\frac{b}{a}$. Flipping $\frac{a}{b}$ to $\frac{b}{a}$ is the reciprocal, and the sign change is the opposite, together the opposite reciprocal. The product test $\frac{a}{b} \cdot \left(-\frac{b}{a}\right) = -1$ falls straight out of this. Understanding the rotation, rather than memorizing the phrase, helps you avoid the frequent error of flipping without changing the sign (which gives a steeper parallel-ish line, not a perpendicular one).

How the SOL examines this topic

• Fill-in-the-blank. Write a line in slope-intercept form from given information and type it.
• Multiple choice. Find a parallel or perpendicular slope, or pick the correct equation.
• Coordinate-plane items. Graph a line from an equation, or build the equation of a drawn line.

Try this

Q1. Write the line with slope $-2$ through $(0, 5)$. [1 point]

• Cue. $b = 5$, so $y = -2x + 5$.

Q2. What slope is parallel to $y = 4x - 1$? [1 point]

• Cue. The same slope, $4$.