3.2 Parallel and Perpendicular Lines

3.2 Parallel and Perpendicular Lines

Quick Answer: Parallel lines have the same slope and different y-intercepts. Perpendicular lines have slopes that are negative reciprocals: if one slope is $m$, the other is $-1/m$ (provided $m$ is defined and nonzero). Write the new line with point-slope form using the given point and the new slope. Horizontal and vertical lines are perpendicular to each other.

The Skills Insight ladder rewards connections among lines, points, and equations: the 237–249 systems band and the 250–262 graph-connection band both assume you can recognize and write parallel and perpendicular lines. AAF does not just ask whether two graphs look parallel — it asks you to produce the equation of the line that meets a geometric condition. This skill sits inside College Board Table 11 Linear applications and graphs (10–15% of AAF, 2–3 CAT items). Drill it on FREE items at /practice/accuplacer-advanced-algebra.

Parallel Lines: Same Slope

Two distinct nonvertical lines are parallel when $m_{\parallel} = m$. They never meet because they rise at the same rate. If they also share a y-intercept they are the same line, not parallel. AAF distractors often reprint the original equation in a different form — $2x - y = 5$ versus $y = 2x - 5$ — and label it a parallel option. Always compare slope and intercept, not the cosmetic arrangement of letters.

Worked example. Write the equation of the line through $(3, -2)$ that is parallel to $y = 4x - 1$.

The given line has slope $4$, so the parallel line has slope $4$ as well. Point-slope:

$$y - (-2) = 4(x - 3)$$ $$y + 2 = 4x - 12$$ $$y = 4x - 14.$$

The two lines are $y = 4x - 1$ and $y = 4x - 14$: identical slopes, intercepts $-1$ and $-14$. They never intersect. Substituting $x = 3$ into the new equation gives $y = 12 - 14 = -2$, so the given point is on the new line.

If the given line is in standard form, solve for $y$ first. Parallel to $3x - 6y = 12$: $-6y = -3x + 12$, so $y = \frac{1}{2}x - 2$. Any parallel line is $y = \frac{1}{2}x + b$ with $b \neq -2$, or, through a specified point, the unique $b$ that fits that point.

Perpendicular Lines: Negative Reciprocal

Two lines with slopes $m_1$ and $m_2$ are perpendicular when $m_1 \cdot m_2 = -1$, equivalently $m_{\perp} = -1/m$. Two operations: take the reciprocal (flip the fraction) and change the sign. Doing only one of those is the most common trap on this skill.

• $m = 2$ $\Rightarrow$ $m_{\perp} = -1/2$ (not $+1/2$, and not $-2$)
• $m = -3/5$ $\Rightarrow$ $m_{\perp} = 5/3$ (flip, and the two negatives cancel)
• $m = -4$ $\Rightarrow$ $m_{\perp} = 1/4$

Worked example. Write the line through $(-1, 5)$ perpendicular to $y = \frac{2}{3}x + 4$.

Given slope $2/3$, perpendicular slope $-3/2$. Point-slope:

$$y - 5 = -\frac{3}{2}(x - (-1)) = -\frac{3}{2}(x + 1)$$ $$y - 5 = -\frac{3}{2}x - \frac{3}{2}$$ $$y = -\frac{3}{2}x + \frac{7}{2}.$$

Check the product of slopes: $(2/3) \cdot (-3/2) = -1$. Check the point: at $x = -1$, $y = -\frac{3}{2}(-1) + \frac{7}{2} = 3/2 + 7/2 = 5$. Both conditions hold, so the line is correct.

Trap: Reciprocal without the Negative

If a stem asks for a perpendicular to $y = 3x + 1$ through $(0, 4)$, the trap answer $y = \frac{1}{3}x + 4$ uses the reciprocal but keeps the original sign. That line is neither parallel (slopes 3 vs $1/3$) nor perpendicular (product $+1$, not $-1$). The correct perpendicular is $y = -\frac{1}{3}x + 4$. A second trap is $y = -3x + 4$, the negative but not the reciprocal: product $-9$, not $-1$. A third trap is $y = -\frac{1}{3}x - 1$, which is perpendicular to the given line but misses the given point.

On a CAT item, scan the options for those three near-misses before you commit. The right choice must pass both tests: slope product $-1$, and the named point satisfies the equation.

Horizontal and Vertical Pairs

A horizontal line ($m = 0$, $y = k$) is perpendicular to every vertical line ($x = h$). A vertical line has undefined slope, so you cannot compute $-1/m$ numerically; you switch families instead.

Worked example. Line through $(4, -3)$ perpendicular to $x = 1$. The given line is vertical, so the perpendicular is horizontal: $y = -3$. The parallel would be another vertical line, $x = 4$.

Never write $m = 0$ and undefined slope as reciprocals of each other in a formula box; the geometric fact is simply that horizontal lines are perpendicular to vertical lines. An item that gives $y = -5$ and asks for a parallel through $(2, 7)$ wants $y = 7$, not $x = 2$.

From Two Conditions to One Equation

Higher-band AAF items stitch three facts together: a point the line must contain, a relationship (parallel or perpendicular) to a given line, and a request for the equation in a specified form. The workflow is always:

1. Extract $m$ from the given line (solve for $y$ if needed).
2. Copy $m$ (parallel) or take the negative reciprocal (perpendicular), watching the horizontal/vertical exception.
3. Plug the given point into point-slope.
4. Convert to the form the item asks for: slope-intercept $y = mx + b$ or standard $Ax + By = C$ with integer coefficients.

Worked example, standard form. Line through $(2, 1)$ parallel to $x + 2y = 8$. Solve: $2y = -x + 8$, $y = -\frac{1}{2}x + 4$, so $m = -1/2$. Then $y - 1 = -\frac{1}{2}(x - 2)$, $y - 1 = -\frac{1}{2}x + 1$, $y = -\frac{1}{2}x + 2$. Multiply by 2: $2y = -x + 4$, so $x + 2y = 4$. Notice this is the same $A$ and $B$ as the original $x + 2y = 8$, with a different $C$ — that is the standard-form signature of parallel lines.

Two lines $A_1 x + B_1 y = C_1$ and $A_2 x + B_2 y = C_2$ are parallel when the ratios $A_1/B_1$ and $A_2/B_2$ match (same slope $-A/B$) and the $C$ values differ. They are perpendicular when $A_1 A_2 + B_1 B_2 = 0$. You do not need vector language on AAF, but that integer check is fast: $x + 2y = 8$ and $2x - y = 3$ give $1 \cdot 2 + 2 \cdot (-1) = 0$, so those two are perpendicular.

After the slope rule, the point check, and the form conversion are automatic, this linear-geometry skill is mostly careful arithmetic. Keep the negative sign glued to the reciprocal, and treat $m = 0$ and undefined slope as a family switch rather than a formula.