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7.1 Advanced Linear Equations, Systems, and Coordinate Geometry

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ACCUPLACER Suite Study Guide Exam

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1 Chapter 1: ACCUPLACER Suite Overview & Testing Mechanics

1.1 The ACCUPLACER Testing System & Computer-Adaptive Mechanics 1.2 Placement Policies, Score Scales & Institutional Cut Scores 1.3 Test-Taking Strategies, Tool Mechanics & Pacing

2 Chapter 2: Next-Generation Reading

2.1 The Reading Test Blueprint: Set-Based and Discrete Items 2.2 Information and Ideas: Close Reading & Central Themes 2.3 Summarizing and Understanding Relationships 2.4 Rhetoric, Text Structure & Author's Purpose 2.5 Analyzing Point of View and Authorial Purpose 2.6 Analyzing Arguments: Claims, Counterclaims, and Evidence 2.7 Synthesis & Paired Text Comparative Analysis 2.8 Vocabulary in Context & Figurative Language 2.9 Passage Genre, Content Area, Mode, and Text Complexity 2.10 Answer-Choice Traps and Evidence Discipline

3 Chapter 3: Next-Generation Writing: Expression of Ideas

3.1 The Writing Test Blueprint: Passage Sets and Content Distribution 3.2 Development: Proposition and Supporting Evidence 3.3 Focus: Adding, Retaining, and Deleting for Relevance 3.4 Organization: Logical Sequence and Paragraph Order 3.5 Introductions, Conclusions, and Transitional Cohesion 3.6 Effective Language Use: Precision and Word Choice 3.7 Concision: Eliminating Wordiness and Redundancy 3.8 Style, Tone, and Voice Consistency 3.9 Syntax: Sentence Combining for Rhetorical Purpose

4 Chapter 4: Next-Generation Writing: Standard English Conventions

4.1 Sentence Boundaries: Fragments, Run-ons, and Comma Splices 4.2 Subordination and Coordination 4.3 Parallel Structure 4.4 Modifier Placement: Misplaced and Dangling Modifiers 4.5 Inappropriate Shifts in Verb Tense 4.6 Inappropriate Shifts in Verb Voice and Mood 4.7 Inappropriate Shifts in Pronoun Person and Number 4.8 Subject-Verb Agreement, Pronoun Case, and Verb Forms 4.9 Possessive Determiners, Contractions, and Confusable Adverbs 4.10 Pronoun Clarity and Pronoun-Antecedent Agreement 4.11 Frequently Confused Words, Noun Agreement, and Logical Comparison 4.12 Within-Sentence Punctuation: Semicolons, Colons, Dashes, and Apostrophes 4.13 Items in a Series and Nonrestrictive Elements 4.14 End-of-Sentence Punctuation, Hyphenation, and Unnecessary Punctuation

5 Chapter 5: Next-Generation Arithmetic

5.1 Whole Number Operations, Estimation & Multi-Step Applications 5.2 Fractions, Mixed Numbers & Operations 5.3 Decimal Operations & Place-Value Conversions 5.4 Percent Calculations, Increase, Decrease, and Applications 5.5 Number Comparisons and Equivalents Across Formats

6 Chapter 6: Next-Generation Quantitative Reasoning, Algebra, and Statistics (QAS)

6.1 Rational Numbers, Absolute Value, and Order of Operations 6.2 Exponents, Radicals, and Scientific Notation 6.3 Ratio, Proportional Relationships, and Unit Conversions 6.4 Creating and Evaluating Algebraic Expressions 6.5 Linear Equations, Inequalities, and Systems of Two Equations 6.6 Linear Applications, Functions, and Graphs 6.7 Probability and Sets 6.8 Descriptive Statistics: Center, Spread, and Graphical Displays 6.9 Geometry for Pre-Algebra and Algebra 1

7 Chapter 7: Next-Generation Advanced Algebra, Quadratics & Polynomials

7.1 Advanced Linear Equations, Systems, and Coordinate Geometry 7.2 Factoring Methods for Quadratics, Cubics, and Polynomials 7.3 Quadratic Equations, Vertex Form & Parabolic Functions 7.4 Polynomial Equations, Roots, and Graphs 7.5 Radical and Rational Equations and Functions 7.6 Geometry for Algebra 1 and Algebra 2

8 Chapter 8: Functions, Exponentials, Logarithms & Trigonometry

8.1 Function Notation, Transformations, Compositions & Inverses 8.2 Domain, Range, and Interpreting Functions in Context 8.3 Exponential & Logarithmic Functions, Laws & Modeling 8.4 Right-Triangle Trigonometry, the Unit Circle, and Radian Measure 8.5 Law of Sines, Law of Cosines, and Trigonometric Equations 8.6 Choosing Your Math Test: Arithmetic, QAS, or AAF

9 Chapter 9: WritePlacer & WritePlacer ESL Essays

9.1 WritePlacer Prompt Format, Planning, and the Writing Process 9.2 The WritePlacer Scoring Rubric & 6 Core Dimensions 9.3 Reading the WritePlacer Score Scale: What Each Band Demonstrates 9.4 Essay Architecture: Thesis, Evidence Integration & Structure 9.5 WritePlacer ESL Scoring Standards & Synthesis

10 Chapter 10: ACCUPLACER ESL Placement Suite

10.1 The ESL Placement Suite: Structure, Scoring, and Score Interpretation 10.2 ESL Reading Skills & Passage Comprehension 10.3 ESL Sentence Meaning & Context Clues 10.4 ESL Language Use & Grammatical Mechanics 10.5 ESL Listening Comprehension & Spoken Contexts

7.1 Advanced Linear Equations, Systems, and Coordinate Geometry

Key Takeaways Explain This Topic AI

Solving 3x3 linear systems requires systematic two-step elimination to reduce three equations in three variables down to a 2x2 system, solving for two variables, and back-substituting to determine the third.

Nonlinear systems (such as line-parabola or line-circle intersections) are solved by algebraic substitution, converting the system into a single quadratic equation where the discriminant (b² - 4ac) determines whether there are 2, 1 (tangent), or 0 real intersection points.

A 2x2 matrix determinant ad - bc governs system solvability: a non-zero determinant guarantees a unique solution, while a determinant of zero signifies either no solution (inconsistent, parallel lines) or infinitely many solutions (dependent, coincident lines).

Coordinate geometry formulas calculate Euclidean distance d = sqrt((x2 - x1)² + (y2 - y1)²), midpoint M = ((x1 + x2)/2, (y1 + y2)/2), and perpendicular bisector lines using negative reciprocal slopes m_perp = -1/m through the segment midpoint.

Circle equations in general form x² + y² + Dx + Ey + F = 0 are converted to standard form (x - h)² + (y - k)² = r² by completing the square separately for x and y terms to identify center (h, k) and radius r.

Last updated: August 2026

Advanced Linear Systems, Coordinate Geometry & Matrices

Core Foundation: The Next-Generation Advanced Algebra and Functions (AAF) assessment measures high-level algebraic reasoning, multi-variable system analysis, and coordinate geometry modeling. Mastery of algebraic elimination in three dimensions, matrix determinants, intersection geometry between linear and conic curves, and completing the square for circular loci is essential for achieving a college-ready placement score.

Systems of Three Linear Equations in Three Variables

A linear equation in three variables ( x , y , z x, y, z x , y , z ) has the standard form A x + B y + C z = D Ax + By + Cz = D A x + B y + C z = D (where A , B , C A, B, C A , B , C are not all zero). Geometrically, each linear equation in three variables represents a flat two-dimensional plane in three-dimensional space ( R 3 \mathbb{R}^3 R 3 ) . +-------------------------------------------------------------------------+

  • | GEOMETRIC INTERPRETATIONS OF 3×3 LINEAR SYSTEMS |
  • +-------------------------------------------------------------------------+
  • | 1. UNIQUE SOLUTION (Consistent & Independent) |
  • | Three planes intersect at exactly ONE single point (x, y, z). |
  • | |
  • | 2. INFINITELY MANY SOLUTIONS (Consistent & Dependent) |
  • | Three planes intersect along a common line or are coincident. |
  • | Algebraic signature: 0 = 0 after elimination. |
  • | |
  • | 3. NO SOLUTION (Inconsistent) |
  • | Planes are parallel or intersect in three pairwise parallel lines |
  • | forming a triangular prism with no shared intersection. |
  • | Algebraic signature: 0 = k (where k ≠ 0) after elimination. |
  • +-------------------------------------------------------------------------+
  • The Systematic Elimination Algorithm for 3×3 Systems

To solve a 3 × 3 3 \times 3 3 × 3 system algebraically without errors, follow this structured reduction protocol:

Select a Target Variable: Choose the variable ( x , y x, y x , y , or z z z ) with the simplest coefficients (ideally coefficients of ± 1 \pm 1 ± 1 ).

Pair the Equations: Use the chosen variable to create two distinct pairs of equations: Pair A (Equations 1 & 2) and Pair B (Equations 2 & 3, or Equations 1 & 3).

Eliminate the Target Variable: Multiply each equation by appropriate constants so that the target variable cancels when the pairs are added, producing a 2 × 2 2 \times 2 2 × 2 system in the remaining two variables.

Solve the 2 × 2 2 \times 2 2 × 2 System: Use elimination or substitution to find the numerical values of the two remaining variables.

Back-Substitute: Substitute the two known values into any of the original three equations to solve for the target variable.

Verify the Ordered Triple ( x , y , z ) (x, y, z) ( x , y , z ) : Check the candidate solution in all three original equations.

Worked Example: Solving a 3×3 System Step-by-Step

Solve the system of equations: ( 1 ) x + 2 y + z = 7 ( 2 ) 2 x − y + 3 z = 12 ( 3 ) 3 x + 3 y − 2 z = 3 \begin{align*}

  • (1) \quad x + 2y + z &= 7 \\
  • (2) \quad 2x - y + 3z &= 12 \\
  • (3) \quad 3x + 3y - 2z &= 3
  • \end{align*} ( 1 ) x + 2 y + z ( 2 ) 2 x − y + 3 z ( 3 ) 3 x + 3 y − 2 z ​ = 7 = 12 = 3 ​

Step 1: Eliminate y y y using Equations (1) and (2). Multiply Equation (2) by 2 2 2 : 2 ( 2 x − y + 3 z ) = 2 ( 12 ) ⟹ 4 x − 2 y + 6 z = 24 2(2x - y + 3z) = 2(12) \implies 4x - 2y + 6z = 24 2 ( 2 x − y + 3 z ) = 2 ( 12 ) ⟹ 4 x − 2 y + 6 z = 24

Add to Equation (1): ( x + 2 y + z ) + ( 4 x − 2 y + 6 z ) = 7 + 24 ⟹ 5 x + 7 z = 31 — (Equation 4) (x + 2y + z) + (4x - 2y + 6z) = 7 + 24 \implies 5x + 7z = 31 \quad \text{--- (Equation 4)} ( x + 2 y + z ) + ( 4 x − 2 y + 6 z ) = 7 + 24 ⟹ 5 x + 7 z = 31 — (Equation 4)

Step 2: Eliminate y y y using Equations (2) and (3). Multiply Equation (2) by 3 3 3 : 3 ( 2 x − y + 3 z ) = 3 ( 12 ) ⟹ 6 x − 3 y + 9 z = 36 3(2x - y + 3z) = 3(12) \implies 6x - 3y + 9z = 36 3 ( 2 x − y + 3 z ) = 3 ( 12 ) ⟹ 6 x − 3 y + 9 z = 36

Add to Equation (3): ( 3 x + 3 y − 2 z ) + ( 6 x − 3 y + 9 z ) = 3 + 36 ⟹ 9 x + 7 z = 39 — (Equation 5) (3x + 3y - 2z) + (6x - 3y + 9z) = 3 + 36 \implies 9x + 7z = 39 \quad \text{--- (Equation 5)} ( 3 x + 3 y − 2 z ) + ( 6 x − 3 y + 9 z ) = 3 + 36 ⟹ 9 x + 7 z = 39 — (Equation 5)

Step 3: Solve the resulting 2 × 2 2 \times 2 2 × 2 system (Equations 4 & 5). Subtract Equation (4) from Equation (5): ( 9 x + 7 z ) − ( 5 x + 7 z ) = 39 − 31 (9x + 7z) - (5x + 7z) = 39 - 31 ( 9 x + 7 z ) − ( 5 x + 7 z ) = 39 − 31 4 x = 8 ⟹ x = 2 4x = 8 \implies x = 2 4 x = 8 ⟹ x = 2

Substitute x = 2 x = 2 x = 2 into Equation (4): 5 ( 2 ) + 7 z = 31 ⟹ 10 + 7 z = 31 ⟹ 7 z = 21 ⟹ z = 3 5(2) + 7z = 31 \implies 10 + 7z = 31 \implies 7z = 21 \implies z = 3 5 ( 2 ) + 7 z = 31 ⟹ 10 + 7 z = 31 ⟹ 7 z = 21 ⟹ z = 3

Step 4: Back-substitute x = 2 x = 2 x = 2 and z = 3 z = 3 z = 3 into Equation (1). 2 + 2 y + 3 = 7 ⟹ 2 y + 5 = 7 ⟹ 2 y = 2 ⟹ y = 1 2 + 2y + 3 = 7 \implies 2y + 5 = 7 \implies 2y = 2 \implies y = 1 2 + 2 y + 3 = 7 ⟹ 2 y + 5 = 7 ⟹ 2 y = 2 ⟹ y = 1

Step 5: Verification in Equation (2) & (3). 2 ( 2 ) − ( 1 ) + 3 ( 3 ) = 4 − 1 + 9 = 12 2(2) - (1) + 3(3) = 4 - 1 + 9 = 12 2 ( 2 ) − ( 1 ) + 3 ( 3 ) = 4 − 1 + 9 = 12 (True); 3 ( 2 ) + 3 ( 1 ) − 2 ( 3 ) = 6 + 3 − 6 = 3 3(2) + 3(1) - 2(3) = 6 + 3 - 6 = 3 3 ( 2 ) + 3 ( 1 ) − 2 ( 3 ) = 6 + 3 − 6 = 3 (True).

The unique solution is the ordered triple ( 2 , 1 , 3 ) (2, 1, 3) ( 2 , 1 , 3 ) .

Nonlinear Systems: Intersections of Lines, Parabolas & Circles

A nonlinear system contains at least one equation of degree two or higher. Solving nonlinear systems on the ACCUPLACER requires substituting linear expressions into quadratic expressions to locate shared coordinate points. +-----------------------------------------------------------------------------------------+

  • | NONLINEAR SYSTEM INTERSECTION PROFILES |
  • +-----------------------+---------------------------------+-------------------------------+
  • | System Configuration | Possible Real Solution Counts | Geometric Relationship |
  • +-----------------------+---------------------------------+-------------------------------+
  • | Line & Parabola | 2 Solutions | Secant line (crosses twice) |
  • | | 1 Solution | Tangent line (touches vertex) |
  • | | 0 Solutions | Disjoint (no intersection) |
  • +-----------------------+---------------------------------+-------------------------------+
  • | Line & Circle | 2 Solutions | Secant line (chords circle) |
  • | | 1 Solution | Tangent line (touches rim) |
  • | | 0 Solutions | Exterior line (no contact) |
  • +-----------------------+---------------------------------+-------------------------------+
  • | Parabola & Parabola | 0, 1, 2, 3, or 4 Solutions | Intersecting conic sections |
  • +-----------------------+---------------------------------+-------------------------------+
  • Line-Parabola Intersection Methodology

Express the linear equation in terms of one variable (e.g., y = m x + b y = mx + b y = m x + b ).

Substitute the linear expression into the quadratic parabola equation y = a x 2 + b x + c y = ax^2 + bx + c y = a x 2 + b x + c .

Set the resulting quadratic equation to zero: a x 2 + ( b − m ) x + ( c − b 0 ) = 0 ax^2 + (b - m)x + (c - b_0) = 0 a x 2 + ( b − m ) x + ( c − b 0 ​ ) = 0 .

Evaluate the discriminant Δ = B 2 − 4 A C \Delta = B^2 - 4AC Δ = B 2 − 4 A C :

Δ > 0 \Delta > 0 Δ > 0 : 2 distinct real intersection points.

Δ = 0 \Delta = 0 Δ = 0 : 1 real intersection point (the line is tangent to the parabola).

Δ < 0 \Delta < 0 Δ < 0 : 0 real intersection points (the curves never cross in the real plane).

Solve for x x x and compute corresponding y y y -values using the linear equation.

Worked Example: Line-Parabola Intersection

Find all intersection points of the system: { y = 2 x − 1 y = x 2 − 3 x + 5 \begin{cases} y = 2x - 1 \\ y = x^2 - 3x + 5 \end{cases} { y = 2 x − 1 y = x 2 − 3 x + 5 ​

Step 1 (Equate expressions): x 2 − 3 x + 5 = 2 x − 1 x^2 - 3x + 5 = 2x - 1 x 2 − 3 x + 5 = 2 x − 1 .

Step 2 (Form standard quadratic): x 2 − 5 x + 6 = 0 x^2 - 5x + 6 = 0 x 2 − 5 x + 6 = 0 .

Step 3 (Factor): ( x − 2 ) ( x − 3 ) = 0 ⟹ x = 2 or x = 3 (x - 2)(x - 3) = 0 \implies x = 2 \quad \text{or} \quad x = 3 ( x − 2 ) ( x − 3 ) = 0 ⟹ x = 2 or x = 3 .

Step 4 (Find corresponding y y y -coordinates):

For x = 2 x = 2 x = 2 : y = 2 ( 2 ) − 1 = 3 ⟹ ( 2 , 3 ) y = 2(2) - 1 = 3 \implies (2, 3) y = 2 ( 2 ) − 1 = 3 ⟹ ( 2 , 3 )

For x = 3 x = 3 x = 3 : y = 2 ( 3 ) − 1 = 5 ⟹ ( 3 , 5 ) y = 2(3) - 1 = 5 \implies (3, 5) y = 2 ( 3 ) − 1 = 5 ⟹ ( 3 , 5 )

The solution set is { ( 2 , 3 ) , ( 3 , 5 ) } \{ (2, 3), (3, 5) \} {( 2 , 3 ) , ( 3 , 5 )} .

Absolute Value Inequalities in Two Variables

An absolute value inequality in two variables represents a two-dimensional shaded region in the Cartesian coordinate plane bounded by a V-shaped graph: y ≥ a ∣ x − h ∣ + k or y < a ∣ x − h ∣ + k y \ge a|x - h| + k \quad \text{or} \quad y < a|x - h| + k y ≥ a ∣ x − h ∣ + k or y < a ∣ x − h ∣ + k

Graphing and Boundary Rules

Vertex and Axis of Symmetry: The boundary vertex is at ( h , k ) (h, k) ( h , k ) . The vertical line x = h x = h x = h is the axis of symmetry.

Slopes of the V-Arms: The right ray has slope + a +a + a ; the left ray has slope − a -a − a . If a < 0 a < 0 a < 0 , the V-shape opens downward.

Boundary Line Style:

Strict inequalities ( < , > <, > < , > ): Draw a dashed boundary line (points on the V are excluded).

Non-strict inequalities ( ≤ , ≥ \le, \ge ≤ , ≥ ): Draw a solid boundary line (points on the V are included).

Shading Region Test:

For y > ∣ x − h ∣ + k y > |x - h| + k y > ∣ x − h ∣ + k or y ≥ ∣ x − h ∣ + k y \ge |x - h| + k y ≥ ∣ x − h ∣ + k : Shade the region above (inside) the V-shape.

For y < ∣ x − h ∣ + k y < |x - h| + k y < ∣ x − h ∣ + k or y ≤ ∣ x − h ∣ + k y \le |x - h| + k y ≤ ∣ x − h ∣ + k : Shade the region below (outside) the V-shape.

Confirm with a test point such as ( 0 , 0 ) (0, 0) ( 0 , 0 ) (provided ( 0 , 0 ) (0,0) ( 0 , 0 ) does not lie directly on the boundary).

Matrix Representations and Determinants for 2×2 Systems

A system of two linear equations in two variables: { a x + b y = e c x + d y = f \begin{cases} ax + by = e \\ cx + dy = f \end{cases} { a x + b y = e c x + d y = f ​

can be written in matrix form as A x = b A\mathbf{x} = \mathbf{b} A x = b : ( a b c d ) ( x y ) = ( e f ) \begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} e \\ f \end{pmatrix} ( a c ​ b d ​ ) ( x y ​ ) = ( e f ​ )

The 2×2 Matrix Determinant

The determinant of the coefficient matrix A A A , denoted det ⁡ ( A ) \det(A) det ( A ) or ∣ A ∣ |A| ∣ A ∣ , is defined as: det ⁡ ( A ) = ∣ a b c d ∣ = a d − b c \det(A) = \begin{vmatrix} a & b \\ c & d \end{vmatrix} = ad - bc det ( A ) = ​ a c ​ b d ​ ​ = a d − b c +-------------------------------------------------------------------------+

  • | DETERMINANT SOLVABILITY CLASSIFICATION |
  • +-----------------------+-----------------------+-------------------------+
  • | Determinant Value | System Nature | Graphical Geometry |
  • +-----------------------+-----------------------+-------------------------+
  • | det(A) = ad - bc ≠ 0 | Unique Solution | Intersecting Lines |
  • | | (Consistent/Indep.) | (Single (x, y) point) |
  • +-----------------------+-----------------------+-------------------------+
  • | det(A) = 0 and | No Solution | Parallel Lines |
  • | (af - ce ≠ 0) | (Inconsistent) | (Never intersect) |
  • +-----------------------+-----------------------+-------------------------+
  • | det(A) = 0 and | Infinitely Many | Coincident Lines |
  • | (af - ce = 0) | (Consistent/Dep.) | (Identical graph line) |
  • +-----------------------+-----------------------+-------------------------+
  • Cramer's Rule for 2×2 Systems

When det ⁡ ( A ) = D = a d − b c ≠ 0 \det(A) = D = ad - bc \neq 0 det ( A ) = D = a d − b c  = 0 , the exact solution is given by: x = D x D = ∣ e b f d ∣ a d − b c = e d − b f a d − b c , y = D y D = ∣ a e c f ∣ a d − b c = a f − c e a d − b c x = \frac{D_x}{D} = \frac{\begin{vmatrix} e & b \\ f & d \end{vmatrix}}{ad - bc} = \frac{ed - bf}{ad - bc}, \qquad y = \frac{D_y}{D} = \frac{\begin{vmatrix} a & e \\ c & f \end{vmatrix}}{ad - bc} = \frac{af - ce}{ad - bc} x = D D x ​ ​ = a d − b c ​ e f ​ b d ​ ​ ​ = a d − b c e d − b f ​ , y = D D y ​ ​ = a d − b c ​ a c ​ e f ​ ​ ​ = a d − b c a f − ce ​

Coordinate Geometry Foundations

Coordinate geometry unifies algebraic equations with Euclidean spatial metrics.

Essential Coordinate Geometry Formulas

Concept / Metric Mathematical Formula Key Application / Notes

Distance Formula d = ( x 2 − x 1 ) 2 + ( y 2 − y 1 ) 2 d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} d = ( x 2 ​ − x 1 ​ ) 2 + ( y 2 ​ − y 1 ​ ) 2 ​ Derived directly from the Pythagorean theorem ( a 2 + b 2 = c 2 a^2 + b^2 = c^2 a 2 + b 2 = c 2 )

Midpoint Formula M = ( x 1 + x 2 2 , y 1 + y 2 2 ) M = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right) M = ( 2 x 1 ​ + x 2 ​ ​ , 2 y 1 ​ + y 2 ​ ​ ) Arithmetic average of x x x -coordinates and y y y -coordinates

Slope of a Line m = y 2 − y 1 x 2 − x 1 ( x 1 ≠ x 2 ) m = \frac{y_2 - y_1}{x_2 - x_1} \quad (x_1 \neq x_2) m = x 2 ​ − x 1 ​ y 2 ​ − y 1 ​ ​ ( x 1 ​  = x 2 ​ ) Vertical change over horizontal change (rise over run)

Parallel Lines Slope m 1 = m 2 m_1 = m_2 m 1 ​ = m 2 ​ Equal slopes with distinct y y y -intercepts

Perpendicular Slope m 1 ⋅ m 2 = − 1 ⟺ m 2 = − 1 m 1 m_1 \cdot m_2 = -1 \iff m_2 = -\frac{1}{m_1} m 1 ​ ⋅ m 2 ​ = − 1 ⟺ m 2 ​ = − m 1 ​ 1 ​ Slopes are negative reciprocals of each other

Standard Form of Circle ( x − h ) 2 + ( y − k ) 2 = r 2 (x - h)^2 + (y - k)^2 = r^2 ( x − h ) 2 + ( y − k ) 2 = r 2 Center ( h , k ) (h, k) ( h , k ) , radius r = r 2 r = \sqrt{r^2} r = r 2 ​

Perpendicular Bisector Protocol

The perpendicular bisector of a line segment connecting points A ( x 1 , y 1 ) A(x_1, y_1) A ( x 1 ​ , y 1 ​ ) and B ( x 2 , y 2 ) B(x_2, y_2) B ( x 2 ​ , y 2 ​ ) is the unique line that is perpendicular to segment A B AB A B and passes through its midpoint.

Find the Midpoint ( M M M ): Compute M = ( x 1 + x 2 2 , y 1 + y 2 2 ) M = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right) M = ( 2 x 1 ​ + x 2 ​ ​ , 2 y 1 ​ + y 2 ​ ​ ) .

Find the Slope of the Segment ( m A B m_{AB} m A B ​ ): Compute m A B = y 2 − y 1 x 2 − x 1 m_{AB} = \frac{y_2 - y_1}{x_2 - x_1} m A B ​ = x 2 ​ − x 1 ​ y 2 ​ − y 1 ​ ​ .

Determine the Perpendicular Slope ( m ⊥ m_\perp m ⊥ ​ ): Compute m ⊥ = − 1 m A B m_\perp = -\frac{1}{m_{AB}} m ⊥ ​ = − m A B ​ 1 ​ .

Form the Linear Equation: Use point-slope form with M ( x m , y m ) M(x_m, y_m) M ( x m ​ , y m ​ ) and m ⊥ m_\perp m ⊥ ​ : y − y m = m ⊥ ( x − x m ) y - y_m = m_\perp(x - x_m) y − y m ​ = m ⊥ ​ ( x − x m ​ )

Worked Example: Equation of a Perpendicular Bisector

Find the equation of the perpendicular bisector of the segment connecting A ( − 2 , 5 ) A(-2, 5) A ( − 2 , 5 ) and B ( 4 , − 3 ) B(4, -3) B ( 4 , − 3 ) .

Step 1 (Midpoint): M = ( − 2 + 4 2 , 5 + ( − 3 ) 2 ) = ( 2 2 , 2 2 ) = ( 1 , 1 ) M = \left(\frac{-2 + 4}{2}, \frac{5 + (-3)}{2}\right) = \left(\frac{2}{2}, \frac{2}{2}\right) = (1, 1) M = ( 2 − 2 + 4 ​ , 2 5 + ( − 3 ) ​ ) = ( 2 2 ​ , 2 2 ​ ) = ( 1 , 1 ) .

Step 2 (Slope of A B AB A B ): m A B = − 3 − 5 4 − ( − 2 ) = − 8 6 = − 4 3 m_{AB} = \frac{-3 - 5}{4 - (-2)} = \frac{-8}{6} = -\frac{4}{3} m A B ​ = 4 − ( − 2 ) − 3 − 5 ​ = 6 − 8 ​ = − 3 4 ​ .

Step 3 (Perpendicular slope): m ⊥ = − 1 − 4 / 3 = + 3 4 m_\perp = -\frac{1}{-4/3} = +\frac{3}{4} m ⊥ ​ = − − 4/3 1 ​ = + 4 3 ​ .

Step 4 (Point-slope equation): y − 1 = 3 4 ( x − 1 ) ⟹ y = 3 4 x − 3 4 + 1 ⟹ y = 3 4 x + 1 4 y - 1 = \frac{3}{4}(x - 1) \implies y = \frac{3}{4}x - \frac{3}{4} + 1 \implies y = \frac{3}{4}x + \frac{1}{4} y − 1 = 4 3 ​ ( x − 1 ) ⟹ y = 4 3 ​ x − 4 3 ​ + 1 ⟹ y = 4 3 ​ x + 4 1 ​ (or 3 x − 4 y = − 1 3x - 4y = -1 3 x − 4 y = − 1 ).

Circles: Standard Form and Completing the Square

The general second-degree equation for a circle in the Cartesian plane is: x 2 + y 2 + D x + E y + F = 0 x^2 + y^2 + Dx + Ey + F = 0 x 2 + y 2 + D x + E y + F = 0

To graph the circle or find its center ( h , k ) (h, k) ( h , k ) and radius r r r , you must rewrite this into standard form: ( x − h ) 2 + ( y − k ) 2 = r 2 (x - h)^2 + (y - k)^2 = r^2 ( x − h ) 2 + ( y − k ) 2 = r 2

Step-by-Step Completing the Square for Circles

Group Terms: Group x x x -terms together, group y y y -terms together, and subtract the constant F F F to the right side: ( x 2 + D x ) + ( y 2 + E y ) = − F (x^2 + Dx) + (y^2 + Ey) = -F ( x 2 + D x ) + ( y 2 + E y ) = − F

Complete the Square for x x x and y y y :

Add ( D 2 ) 2 \left(\frac{D}{2}\right)^2 ( 2 D ​ ) 2 inside the x x x -group and to the right side.

Add ( E 2 ) 2 \left(\frac{E}{2}\right)^2 ( 2 E ​ ) 2 inside the y y y -group and to the right side. ( x 2 + D x + ( D / 2 ) 2 ) + ( y 2 + E y + ( E / 2 ) 2 ) = − F + ( D / 2 ) 2 + ( E / 2 ) 2 (x^2 + Dx + (D/2)^2) + (y^2 + Ey + (E/2)^2) = -F + (D/2)^2 + (E/2)^2 ( x 2 + D x + ( D /2 ) 2 ) + ( y 2 + E y + ( E /2 ) 2 ) = − F + ( D /2 ) 2 + ( E /2 ) 2

Factor into Binomial Squares: ( x + D 2 ) 2 + ( y + E 2 ) 2 = r 2 \left(x + \frac{D}{2}\right)^2 + \left(y + \frac{E}{2}\right)^2 = r^2 ( x + 2 D ​ ) 2 + ( y + 2 E ​ ) 2 = r 2

Extract Center and Radius: Center is ( − D 2 , − E 2 ) \left(-\frac{D}{2}, -\frac{E}{2}\right) ( − 2 D ​ , − 2 E ​ ) and radius is r = r 2 r = \sqrt{r^2} r = r 2 ​ (where r 2 > 0 r^2 > 0 r 2 > 0 ).

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