Study guide
3.2 Parallel and Perpendicular Lines
ACCUPLACER Advanced Algebra and Functions Exam
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1 Introduction & Exam Overview
1.1 Current Exam Facts 1.2 Scoring, Skills Insight Bands, and Placement 1.3 How to Study for AAF
2 Linear Equations
2.1 Writing Linear Equations in One and Two Variables 2.2 Solving Linear Equations 2.3 Linear Inequalities 2.4 Systems of Two Linear Equations
3 Linear Applications and Graphs
3.1 Slope, Intercepts, and Graphing Lines 3.2 Parallel and Perpendicular Lines 3.3 Modeling with Linear Functions 3.4 Graphing Linear Inequalities and Systems
4 Factoring
4.1 GCF, Grouping, and Special Products 4.2 Factoring Quadratic Trinomials 4.3 Factoring Cubics and Higher Polynomials
5 Quadratics
5.1 Quadratic Forms, Graphs, Vertex, and Intercepts 5.2 Solving Quadratics by Factoring 5.3 Quadratic Formula and Completing the Square 5.4 Quadratic Inequalities and Quadratic-Linear Systems
6 Functions
6.1 Function Notation, Evaluation, Domain, and Range 6.2 Graphs of Functions, Vertical Line Test, Maxima and Minima 6.3 Function Operations, Composition, and Inverses 6.4 Transformations and Interpreting Functions in Context
7 Radical and Rational Equations
7.1 Simplifying Radical Expressions 7.2 Solving Radical Equations and Extraneous Roots 7.3 Rational Expressions and Equations 7.4 Graphs, Domain, and Range of Radical and Rational Functions
8 Polynomial Equations
8.1 Polynomial Operations and Structure 8.2 Solving Polynomial Equations 8.3 Graphing Polynomials, Roots, and End Behavior
9 Exponential and Logarithmic Equations
9.1 Exponential Functions, Growth, and Decay 9.2 Graphs of Exponential Functions 9.3 Logarithms and Logarithmic Properties 9.4 Solving Exponential and Logarithmic Equations
10 Geometry Concepts for Algebra 1
10.1 Area, Perimeter, and Prism Volume 10.2 Distance Formula and Pythagorean Theorem 10.3 Dilations, Rotations, Translations, and Reflections
11 Geometry Concepts for Algebra 2
11.1 Volume of Cylinders, Cones, and Spheres 11.2 Intersecting Lines and Angle Relationships 11.3 Triangle Similarity and Congruence 11.4 Circles in the Coordinate Plane
12 Trigonometry
12.1 Right-Triangle Trigonometry and Special Triangles 12.2 Unit Circle, Radians, and Arc Length 12.3 Trigonometric Graphs and Identities 12.4 Solving Trig Equations; Law of Sines and Cosines
3.2 Parallel and Perpendicular Lines
Key Takeaways Explain This Topic AI
Parallel nonvertical lines satisfy m_parallel = m and have different y-intercepts; the same slope and the same intercept means the same line, not a parallel pair.
Perpendicular slopes are negative reciprocals: m_perp = −1/m when m is defined and nonzero, equivalently m1 · m2 = −1.
The line through (3, −2) parallel to y = 4x − 1 is y = 4x − 14; the line through (−1, 5) perpendicular to y = (2/3)x + 4 is y = −(3/2)x + 7/2.
A horizontal line y = k is perpendicular to every vertical line x = h; do not try to compute −1/m when slope is 0 or undefined.
Writing the parallel or perpendicular line through a given point combines the slope rules with point-slope form; AAF tests the combination, not the rules in isolation.
Last updated: August 2026
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 m m , the other is − 1 / m -1/m − 1/ m (provided m m 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 ∥ = m m_{\parallel} = m m ∥ = 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 — 2 x − y = 5 2x - y = 5 2 x − y = 5 versus y = 2 x − 5 y = 2x - 5 y = 2 x − 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 ) (3, -2) ( 3 , − 2 ) that is parallel to y = 4 x − 1 y = 4x - 1 y = 4 x − 1 .
The given line has slope 4 4 4 , so the parallel line has slope 4 4 4 as well. Point-slope: y − ( − 2 ) = 4 ( x − 3 ) y - (-2) = 4(x - 3) y − ( − 2 ) = 4 ( x − 3 ) y + 2 = 4 x − 12 y + 2 = 4x - 12 y + 2 = 4 x − 12 y = 4 x − 14. y = 4x - 14. y = 4 x − 14.
The two lines are y = 4 x − 1 y = 4x - 1 y = 4 x − 1 and y = 4 x − 14 y = 4x - 14 y = 4 x − 14 : identical slopes, intercepts − 1 -1 − 1 and − 14 -14 − 14 . They never intersect. Substituting x = 3 x = 3 x = 3 into the new equation gives y = 12 − 14 = − 2 y = 12 - 14 = -2 y = 12 − 14 = − 2 , so the given point is on the new line.
If the given line is in standard form, solve for y y y first. Parallel to 3 x − 6 y = 12 3x - 6y = 12 3 x − 6 y = 12 : − 6 y = − 3 x + 12 -6y = -3x + 12 − 6 y = − 3 x + 12 , so y = 1 2 x − 2 y = \frac{1}{2}x - 2 y = 2 1 x − 2 . Any parallel line is y = 1 2 x + b y = \frac{1}{2}x + b y = 2 1 x + b with b ≠ − 2 b \neq -2 b = − 2 , or, through a specified point, the unique b b b that fits that point.
Perpendicular Lines: Negative Reciprocal
Two lines with slopes m 1 m_1 m 1 and m 2 m_2 m 2 are perpendicular when m 1 ⋅ m 2 = − 1 m_1 \cdot m_2 = -1 m 1 ⋅ m 2 = − 1 , equivalently m ⊥ = − 1 / m m_{\perp} = -1/m m ⊥ = − 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 m = 2 m = 2 ⇒ \Rightarrow ⇒ m ⊥ = − 1 / 2 m_{\perp} = -1/2 m ⊥ = − 1/2 (not + 1 / 2 +1/2 + 1/2 , and not − 2 -2 − 2 )
m = − 3 / 5 m = -3/5 m = − 3/5 ⇒ \Rightarrow ⇒ m ⊥ = 5 / 3 m_{\perp} = 5/3 m ⊥ = 5/3 (flip, and the two negatives cancel)
m = − 4 m = -4 m = − 4 ⇒ \Rightarrow ⇒ m ⊥ = 1 / 4 m_{\perp} = 1/4 m ⊥ = 1/4
Worked example. Write the line through ( − 1 , 5 ) (-1, 5) ( − 1 , 5 ) perpendicular to y = 2 3 x + 4 y = \frac{2}{3}x + 4 y = 3 2 x + 4 .
Given slope 2 / 3 2/3 2/3 , perpendicular slope − 3 / 2 -3/2 − 3/2 . Point-slope: y − 5 = − 3 2 ( x − ( − 1 ) ) = − 3 2 ( x + 1 ) y - 5 = -\frac{3}{2}(x - (-1)) = -\frac{3}{2}(x + 1) y − 5 = − 2 3 ( x − ( − 1 )) = − 2 3 ( x + 1 ) y − 5 = − 3 2 x − 3 2 y - 5 = -\frac{3}{2}x - \frac{3}{2} y − 5 = − 2 3 x − 2 3 y = − 3 2 x + 7 2 . y = -\frac{3}{2}x + \frac{7}{2}. y = − 2 3 x + 2 7 .
Check the product of slopes: ( 2 / 3 ) ⋅ ( − 3 / 2 ) = − 1 (2/3) \cdot (-3/2) = -1 ( 2/3 ) ⋅ ( − 3/2 ) = − 1 . Check the point: at x = − 1 x = -1 x = − 1 , y = − 3 2 ( − 1 ) + 7 2 = 3 / 2 + 7 / 2 = 5 y = -\frac{3}{2}(-1) + \frac{7}{2} = 3/2 + 7/2 = 5 y = − 2 3 ( − 1 ) + 2 7 = 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 = 3 x + 1 y = 3x + 1 y = 3 x + 1 through ( 0 , 4 ) (0, 4) ( 0 , 4 ) , the trap answer y = 1 3 x + 4 y = \frac{1}{3}x + 4 y = 3 1 x + 4 uses the reciprocal but keeps the original sign. That line is neither parallel (slopes 3 vs 1 / 3 1/3 1/3 ) nor perpendicular (product + 1 +1 + 1 , not − 1 -1 − 1 ). The correct perpendicular is y = − 1 3 x + 4 y = -\frac{1}{3}x + 4 y = − 3 1 x + 4 . A second trap is y = − 3 x + 4 y = -3x + 4 y = − 3 x + 4 , the negative but not the reciprocal: product − 9 -9 − 9 , not − 1 -1 − 1 . A third trap is y = − 1 3 x − 1 y = -\frac{1}{3}x - 1 y = − 3 1 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 -1 − 1 , and the named point satisfies the equation.
Horizontal and Vertical Pairs
A horizontal line ( m = 0 m = 0 m = 0 , y = k y = k y = k ) is perpendicular to every vertical line ( x = h x = h x = h ). A vertical line has undefined slope, so you cannot compute − 1 / m -1/m − 1/ m numerically; you switch families instead.
Given line Parallel through ( x 0 , y 0 ) (x_0, y_0) ( x 0 , y 0 ) Perpendicular through ( x 0 , y 0 ) (x_0, y_0) ( x 0 , y 0 )
- y = 5 y = 5 y = 5 (horizontal) y = y 0 y = y_0 y = y 0 x = x 0 x = x_0 x = x 0
- x = − 2 x = -2 x = − 2 (vertical) x = x 0 x = x_0 x = x 0 y = y 0 y = y_0 y = y 0
y = 2 x + 1 y = 2x + 1 y = 2 x + 1 y − y 0 = 2 ( x − x 0 ) y - y_0 = 2(x - x_0) y − y 0 = 2 ( x − x 0 ) y − y 0 = − 1 2 ( x − x 0 ) y - y_0 = -\frac{1}{2}(x - x_0) y − y 0 = − 2 1 ( x − x 0 )
Worked example. Line through ( 4 , − 3 ) (4, -3) ( 4 , − 3 ) perpendicular to x = 1 x = 1 x = 1 . The given line is vertical, so the perpendicular is horizontal: y = − 3 y = -3 y = − 3 . The parallel would be another vertical line, x = 4 x = 4 x = 4 .
Never write m = 0 m = 0 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 y = -5 y = − 5 and asks for a parallel through ( 2 , 7 ) (2, 7) ( 2 , 7 ) wants y = 7 y = 7 y = 7 , not x = 2 x = 2 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:
Extract m m m from the given line (solve for y y y if needed).
Copy m m m (parallel) or take the negative reciprocal (perpendicular), watching the horizontal/vertical exception.
Plug the given point into point-slope.
Convert to the form the item asks for: slope-intercept y = m x + b y = mx + b y = m x + b or standard A x + B y = C Ax + By = C A x + B y = C with integer coefficients.
Worked example, standard form. Line through ( 2 , 1 ) (2, 1) ( 2 , 1 ) parallel to x + 2 y = 8 x + 2y = 8 x + 2 y = 8 . Solve: 2 y = − x + 8 2y = -x + 8 2 y = − x + 8 , y = − 1 2 x + 4 y = -\frac{1}{2}x + 4 y = − 2 1 x + 4 , so m = − 1 / 2 m = -1/2 m = − 1/2 . Then y − 1 = − 1 2 ( x − 2 ) y - 1 = -\frac{1}{2}(x - 2) y − 1 = − 2 1 ( x − 2 ) , y − 1 = − 1 2 x + 1 y - 1 = -\frac{1}{2}x + 1 y − 1 = − 2 1 x + 1 , y = − 1 2 x + 2 y = -\frac{1}{2}x + 2 y = − 2 1 x + 2 . Multiply by 2: 2 y = − x + 4 2y = -x + 4 2 y = − x + 4 , so x + 2 y = 4 x + 2y = 4 x + 2 y = 4 . Notice this is the same A A A and B B B as the original x + 2 y = 8 x + 2y = 8 x + 2 y = 8 , with a different C C C — that is the standard-form signature of parallel lines.
Two lines A 1 x + B 1 y = C 1 A_1 x + B_1 y = C_1 A 1 x + B 1 y = C 1 and A 2 x + B 2 y = C 2 A_2 x + B_2 y = C_2 A 2 x + B 2 y = C 2 are parallel when the ratios A 1 / B 1 A_1/B_1 A 1 / B 1 and A 2 / B 2 A_2/B_2 A 2 / B 2 match (same slope − A / B -A/B − A / B ) and the C C C values differ. They are perpendicular when A 1 A 2 + B 1 B 2 = 0 A_1 A_2 + B_1 B_2 = 0 A 1 A 2 + B 1 B 2 = 0 . You do not need vector language on AAF, but that integer check is fast: x + 2 y = 8 x + 2y = 8 x + 2 y = 8 and 2 x − y = 3 2x - y = 3 2 x − y = 3 give 1 ⋅ 2 + 2 ⋅ ( − 1 ) = 0 1 \cdot 2 + 2 \cdot (-1) = 0 1 ⋅ 2 + 2 ⋅ ( − 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 m = 0 m = 0 and undefined slope as a family switch rather than a formula.