Graphing Linear Equations

Turning an Equation into a Line

Every linear equation describes a straight line in the $xy$-plane, and the SAT expects you to move confidently between the algebra and the picture. Graphing a line comes down to finding two reliable points and connecting them — or reading the slope and $y$-intercept directly from the equation. Both approaches appear on the test, and knowing when to use each saves time.

Intercepts: The Two Easiest Points

The $x$-intercept is where the line crosses the horizontal axis, which happens when $y = 0$. The $y$-intercept is where the line crosses the vertical axis, which happens when $x = 0$. Because one coordinate is zero, intercepts are usually the fastest points to compute, especially from standard form $Ax + By = C$.

Take $2x + 3y = 12$. Set $y = 0$ to find the $x$-intercept:

$2x = 12 \;\Rightarrow\; x = 6 \quad\Rightarrow\quad (6, 0)$

Set $x = 0$ to find the $y$-intercept:

$3y = 12 \;\Rightarrow\; y = 4 \quad\Rightarrow\quad (0, 4)$

Plot those two points and draw the line — no slope calculation required. The diagram beside this lesson shows the general picture: a slanted line meets each axis exactly once.

Using Slope and Intercept

When an equation is already in slope-intercept form $y = mx + b$, graphing is even faster. Plot the $y$-intercept $(0, b)$ first, then use the slope as rise over run to step to a second point. A slope of $-\tfrac{2}{3}$ means "down 2, right 3" from the intercept. Connect the two points and extend the line in both directions.

If an equation is in standard form and you prefer the slope method, convert first. For $4x - 2y = 8$:

$\begin{aligned} 4x - 2y &= 8 \\ -2y &= -4x + 8 \\ y &= 2x - 4 \end{aligned}$

Now the $y$-intercept is $-4$, so the line passes through $(0, -4)$; the slope $2 = \tfrac{2}{1}$ means "up 2, right 1," landing on $(1, -2)$ and then $(2, 0)$. That last point is the $x$-intercept, a nice self-check.

Horizontal and Vertical Lines

Two special forms trip students up:

• $y = k$ (a constant) is a horizontal line with slope $0$, crossing the $y$-axis at $k$.
• $x = h$ (a constant) is a vertical line with undefined slope, crossing the $x$-axis at $h$.

A vertical line $x = h$ is not a function — it fails the vertical-line test — yet it is still a valid line the SAT may ask you to graph or interpret. Do not call its slope zero; a run of zero makes the slope undefined.

Parallel and Perpendicular Lines in the Plane

Graphing questions frequently ask you to relate two lines. Parallel lines never intersect and share the same slope. Perpendicular lines meet at a right angle, and their slopes are negative reciprocals whose product is $-1$.

Suppose line $p$ is $y = -3x + 5$ and you need a line parallel to $p$ through $(2, 1)$. Parallel means the slope stays $-3$; use point-slope form:

$\begin{aligned} y - 1 &= -3(x - 2) \\ y - 1 &= -3x + 6 \\ y &= -3x + 7 \end{aligned}$

The parallel line is $y = -3x + 7$. Notice the slope matches $p$ but the $y$-intercept is different ($7$ versus $5$) — that is exactly what "parallel" requires. If instead you needed a perpendicular line, you would swap the slope to $\tfrac{1}{3}$ (the negative reciprocal of $-3$) and repeat the same point-slope process.

Reading a Graph Backward

The SAT also gives you a graph and asks for the equation, or gives two points and asks where the line crosses an axis. The method is the same: get the slope, build the equation with point-slope, then set $y = 0$ (for the $x$-intercept) or read off $b$ (for the $y$-intercept). For a line through $(-2, 7)$ and $(4, -5)$:

$m = \frac{-5 - 7}{4 - (-2)} = \frac{-12}{6} = -2, \qquad y = -2x + 3$

Setting $y = 0$ gives $x = \tfrac{3}{2}$, so it crosses the $x$-axis at $\left(\tfrac{3}{2}, 0\right)$.

Habits That Prevent Errors

Keep the intercept rule straight — $x$-intercept from $y = 0$, $y$-intercept from $x = 0$ — never reverse rise and run when stepping off a slope, and remember that $x = 3$ is a vertical line, not a horizontal one. These three habits eliminate most graphing mistakes on the test.