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2.5 Coordinate Geometry, Lines, and Functions

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Adamson University College Entrance Test AdUCET

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1 Introduction & Exam Overview

1.1 AdUCET Overview, Format, and Admission Process 1.2 Test-Taking and Time-Management Strategies

2 Chapter 2: Mathematics I — Numbers, Algebra, and Functions

2.1 Number Sense, Operations, and Fraction–Decimal–Percent Conversions 2.2 Algebra, Polynomials, and Linear/Quadratic Equations 2.3 Inequalities, Absolute Value, and Systems of Equations 2.4 Radicals, Rational Expressions, and Scientific Notation 2.5 Coordinate Geometry, Lines, and Functions

3 Chapter 3: Mathematics II — Geometry, Measurement, and Trigonometry

3.1 Plane Geometry: Angles, Triangles, Polygons, and Circles 3.2 Solid Mensuration, Measurement, and Unit Conversion 3.3 Right-Triangle Trigonometry and Applications

4 Chapter 4: Mathematics III — Sequences, Proportion, and Data

4.1 Arithmetic and Geometric Sequences and Series 4.2 Ratio, Proportion, Percentage, Work, and Mixture Problems 4.3 Descriptive Statistics and Data Interpretation 4.4 Probability: Rules, Compound Events, and Conditional Probability 4.5 Counting Principles: Permutations and Combinations

5 Chapter 5: English I — Reading Comprehension and Vocabulary

5.1 Reading Comprehension: Main Idea, Purpose, and Passage Structure 5.2 Inference, Supporting Detail, and Author's Tone 5.3 Word Roots, Prefixes, and Suffixes 5.4 Vocabulary in Context, Synonyms, Antonyms, and Confusables

6 Chapter 6: English II — Grammar, Usage, and Written Expression

6.1 Subject-Verb Agreement and Verb Tenses 6.2 Pronouns, Modifiers, and Parallel Structure 6.3 Sentence Boundaries, Clauses, and Punctuation 6.4 Sentence Correction, Redundancy, and Idiomatic Usage 6.5 Verbal Analogies and Relationship Bridges 6.6 Paragraph Organization, Coherence, and Transitions

7 Chapter 7: General Science I — Biology and Chemistry

7.1 Cell Biology, Transport, and Cellular Energetics 7.2 Genetics, Heredity, and Molecular Biology 7.3 Human Body Systems, Ecology, and Evolution 7.4 Matter, Phase Changes, and Atomic Structure 7.5 Periodic Trends, Chemical Bonding, Reactions, Acids and Bases

8 Chapter 8: General Science II — Physics and Earth Science

8.1 Mechanics: Motion, Newton's Laws, Work, and Energy 8.2 Waves, Sound, Light, Heat, Electricity, and Magnetism 8.3 Earth's Structure, Plate Tectonics, and the Rock Cycle 8.4 The Atmosphere, Weather, the Water Cycle, and Astronomy

9 Chapter 9: Aptitude and Abstract Reasoning

9.1 Figure Series and Pattern Progression 9.2 Matrix Reasoning and Figural Analogies 9.3 Spatial Reasoning: Rotation, Reflection, and Chirality 9.4 Paper Folding, Cube Nets, and Orthographic Views 9.5 Deductive Reasoning, Syllogisms, and Venn Diagrams 9.6 Number, Letter, and Alphanumeric Series 9.7 Odd-One-Out, Classification, and Logic Puzzles

2.5 Coordinate Geometry, Lines, and Functions

Key Takeaways Explain This Topic AI

The 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 ​ and 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 ​ ​ ) measure Euclidean line segment properties in the Cartesian plane.

Parallel lines possess identical slopes ( m 1 = m 2 m_1 = m_2 m 1 ​ = m 2 ​ ), whereas perpendicular lines have slopes that are negative reciprocals ( m 1 ⋅ m 2 = − 1 m_1 \cdot m_2 = -1 m 1 ​ ⋅ m 2 ​ = − 1 ).

The slope-intercept form y = m x + b y = mx + b y = m x + b highlights slope m m m and y y y -intercept b b b , while standard form A x + B y = C Ax + By = C A x + B y = C allows rapid calculation of x x x -intercept ( y = 0 y=0 y = 0 ) and y y y -intercept ( x = 0 x=0 x = 0 ).

The domain of a real-valued function excludes values that cause division by zero or negative radicands under even roots.

Composite functions ( f ∘ g ) ( x ) = f ( g ( x ) ) (f \circ g)(x) = f(g(x)) ( f ∘ g ) ( x ) = f ( g ( x )) evaluate the inner function g ( x ) g(x) g ( x ) first before passing its output into the outer function f ( x ) f(x) f ( x ) .

Last updated: August 2026

Coordinate geometry connects algebraic equations with geometric visualizations on the two-dimensional Cartesian coordinate plane. Understanding lines, slopes, distances, function mappings, domain/range constraints, and function compositions is essential for solving spatial and analytical quantitative problems on the AdUCET.

1. The Cartesian Plane and Fundamental Formulas

The Cartesian coordinate plane is formed by two perpendicular axes intersecting at the origin ( 0 , 0 ) (0,0) ( 0 , 0 ) : the horizontal x x x -axis and the vertical y y y -axis. Any point in the plane is represented by an ordered pair ( x , y ) (x, y) ( x , y ) .

The Distance Formula

Derived directly from the Pythagorean Theorem ( a 2 + b 2 = c 2 a^2 + b^2 = c^2 a 2 + b 2 = c 2 ), the distance d d d between two points P 1 ( x 1 , y 1 ) P_1(x_1, y_1) P 1 ​ ( x 1 ​ , y 1 ​ ) and P 2 ( x 2 , y 2 ) P_2(x_2, y_2) P 2 ​ ( x 2 ​ , y 2 ​ ) is: 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 ​

The Midpoint Formula

The midpoint M M M of the line segment connecting P 1 ( x 1 , y 1 ) P_1(x_1, y_1) P 1 ​ ( x 1 ​ , y 1 ​ ) and P 2 ( x 2 , y 2 ) P_2(x_2, y_2) P 2 ​ ( x 2 ​ , y 2 ​ ) represents the average of their respective x x x - and y y y -coordinates: 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 ​ ​ )

Worked Step-by-Step Example

Find the distance d d d and midpoint M M M between points A ( − 2 , 5 ) A(-2, 5) A ( − 2 , 5 ) and B ( 4 , − 3 ) B(4, -3) B ( 4 , − 3 ) .

Distance Calculation: d = ( 4 − ( − 2 ) ) 2 + ( − 3 − 5 ) 2 = ( 6 ) 2 + ( − 8 ) 2 = 36 + 64 = 100 = 10 d = \sqrt{(4 - (-2))^2 + (-3 - 5)^2} = \sqrt{(6)^2 + (-8)^2} = \sqrt{36 + 64} = \sqrt{100} = 10 d = ( 4 − ( − 2 ) ) 2 + ( − 3 − 5 ) 2 ​ = ( 6 ) 2 + ( − 8 ) 2 ​ = 36 + 64 ​ = 100 ​ = 10

Midpoint Calculation: 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 )

2. Slope of a Line

The slope m m m measures the steepness and direction of a line, defined as the ratio of vertical change ("rise") to horizontal change ("run") between any two distinct points ( x 1 , y 1 ) (x_1, y_1) ( x 1 ​ , y 1 ​ ) and ( x 2 , y 2 ) (x_2, y_2) ( x 2 ​ , y 2 ​ ) : m = Δ y Δ x = y 2 − y 1 x 2 − x 1 ( x 1 ≠ x 2 ) m = \frac{\Delta y}{\Delta x} = \frac{y_2 - y_1}{x_2 - x_1} \quad (x_1 \neq x_2) m = Δ x Δ y ​ = x 2 ​ − x 1 ​ y 2 ​ − y 1 ​ ​ ( x 1 ​  = x 2 ​ )

Four Slope Orientations

  • Slope Value Line Orientation Geometric Description
  • m > 0 m > 0 m > 0 Rises from left to right Positive rate of change
  • m < 0 m < 0 m < 0 Falls from left to right Negative rate of change
  • m = 0 m = 0 m = 0 Horizontal line Zero change in y y y ; equation y = c y = c y = c
  • Undefined Vertical line Zero change in x x x ( Δ x = 0 \Delta x = 0 Δ x = 0 ); equation x = c x = c x = c
  • 3. Standard Forms of Linear Equations

Linear equations can be expressed in three main algebraic formats:

Format Name Equation Key Features & Applications

Slope-Intercept Form y = m x + b y = mx + b y = m x + b m m m is the slope; b b b is the y y y -intercept ( 0 , b ) (0, b) ( 0 , b ) . Ideal for graphing.

Point-Slope Form y − y 1 = m ( x − x 1 ) y - y_1 = m(x - x_1) y − y 1 ​ = m ( x − x 1 ​ ) Requires slope m m m and one known point ( x 1 , y 1 ) (x_1, y_1) ( x 1 ​ , y 1 ​ ) . Useful for constructing line equations.

Standard Form A x + B y = C Ax + By = C A x + B y = C A , B , C A, B, C A , B , C are integers ( A ≥ 0 A \ge 0 A ≥ 0 ). Convenient for finding x x x - and y y y -intercepts.

Finding Line Intercepts

x x x -intercept: Set y = 0 y = 0 y = 0 and solve for x x x . The line crosses the horizontal axis at ( x int , 0 ) (x_{\text{int}}, 0) ( x int ​ , 0 ) .

y y y -intercept: Set x = 0 x = 0 x = 0 and solve for y y y . The line crosses the vertical axis at ( 0 , y int ) (0, y_{\text{int}}) ( 0 , y int ​ ) .

Worked Example: Intercepts from Standard Form

Given 3 x − 4 y = 12 3x - 4y = 12 3 x − 4 y = 12 :

To find x x x -intercept: 3 x − 4 ( 0 ) = 12 ⟹ 3 x = 12 ⟹ x = 4 3x - 4(0) = 12 \implies 3x = 12 \implies x = 4 3 x − 4 ( 0 ) = 12 ⟹ 3 x = 12 ⟹ x = 4 . Point is ( 4 , 0 ) (4,0) ( 4 , 0 ) .

To find y y y -intercept: 3 ( 0 ) − 4 y = 12 ⟹ − 4 y = 12 ⟹ y = − 3 3(0) - 4y = 12 \implies -4y = 12 \implies y = -3 3 ( 0 ) − 4 y = 12 ⟹ − 4 y = 12 ⟹ y = − 3 . Point is ( 0 , − 3 ) (0,-3) ( 0 , − 3 ) .

4. Parallel and Perpendicular Lines

The geometric relationship between two lines on the Cartesian plane is dictated entirely by their slopes m 1 m_1 m 1 ​ and m 2 m_2 m 2 ​ .

Parallel Lines ( m 1 = m 2 m_1 = m_2 m 1 ​ = m 2 ​ )

Parallel lines run infinitely in the same direction without intersecting. They have identical slopes and distinct y y y -intercepts: m 1 = m 2 ( b 1 ≠ b 2 ) m_1 = m_2 \quad (b_1 \neq b_2) m 1 ​ = m 2 ​ ( b 1 ​  = b 2 ​ )

Perpendicular Lines ( m 1 ⋅ m 2 = − 1 m_1 \cdot m_2 = -1 m 1 ​ ⋅ m 2 ​ = − 1 )

Perpendicular lines intersect at right angles ( 90 ∘ 90^\circ 9 0 ∘ ). Their slopes are negative reciprocals of each other: m 2 = − 1 m 1 ⟹ m 1 ⋅ m 2 = − 1 m_2 = -\frac{1}{m_1} \implies m_1 \cdot m_2 = -1 m 2 ​ = − m 1 ​ 1 ​ ⟹ m 1 ​ ⋅ m 2 ​ = − 1

Worked Example: Constructing a Perpendicular Line

Find the equation of the line passing through point P ( 3 , − 2 ) P(3, -2) P ( 3 , − 2 ) that is perpendicular to the line y = 1 2 x + 4 y = \frac{1}{2}x + 4 y = 2 1 ​ x + 4 .

Identify Given Slope: The given line has slope m 1 = 1 2 m_1 = \frac{1}{2} m 1 ​ = 2 1 ​ .

Determine Perpendicular Slope: m 2 = − 1 1 / 2 = − 2 m_2 = -\frac{1}{1/2} = -2 m 2 ​ = − 1/2 1 ​ = − 2 .

Apply Point-Slope Form: y − ( − 2 ) = − 2 ( x − 3 ) ⟹ y + 2 = − 2 x + 6 ⟹ y = − 2 x + 4 y - (-2) = -2(x - 3) \implies y + 2 = -2x + 6 \implies y = -2x + 4 y − ( − 2 ) = − 2 ( x − 3 ) ⟹ y + 2 = − 2 x + 6 ⟹ y = − 2 x + 4

5. Functions, Domain, and Range

A function f f f is a binary relation assigning to each element x x x in a set called the domain exactly one element y = f ( x ) y = f(x) y = f ( x ) in a set called the range .

The Vertical Line Test

A graph in the coordinate plane represents a function if and only if no vertical line intersects the curve at more than one point.

Domain Restrictions

When determining the implicit domain of a real-valued algebraic function, apply two main rules:

Division by Zero: Denominators cannot equal zero.

Example: For f ( x ) = 5 x − 3 f(x) = \frac{5}{x - 3} f ( x ) = x − 3 5 ​ , set x − 3 ≠ 0 ⟹ Domain: x ≠ 3 x - 3 \neq 0 \implies \text{Domain: } x \neq 3 x − 3  = 0 ⟹ Domain: x  = 3 or ( − ∞ , 3 ) ∪ ( 3 , ∞ ) (-\infty, 3) \cup (3, \infty) ( − ∞ , 3 ) ∪ ( 3 , ∞ ) .

Even Radicands: Expressions inside even roots (e.g., square roots) must be non-negative.

Example: For g ( x ) = 2 x − 8 g(x) = \sqrt{2x - 8} g ( x ) = 2 x − 8 ​ , set 2 x − 8 ≥ 0 ⟹ 2 x ≥ 8 ⟹ Domain: x ≥ 4 2x - 8 \ge 0 \implies 2x \ge 8 \implies \text{Domain: } x \ge 4 2 x − 8 ≥ 0 ⟹ 2 x ≥ 8 ⟹ Domain: x ≥ 4 or [ 4 , ∞ ) [4, \infty) [ 4 , ∞ ) .

Determining Function Range

The range consists of all possible output values y y y produced by f ( x ) f(x) f ( x ) across its domain:

Linear Functions ( f ( x ) = m x + b , m ≠ 0 f(x) = mx + b, m \neq 0 f ( x ) = m x + b , m  = 0 ): Range is all real numbers ( − ∞ , ∞ ) (-\infty, \infty) ( − ∞ , ∞ ) .

Quadratic Functions ( f ( x ) = a x 2 + b x + c f(x) = ax^2 + bx + c f ( x ) = a x 2 + b x + c ): Range depends on the vertex y y y -coordinate k = f ( − b 2 a ) k = f\left(-\frac{b}{2a}\right) k = f ( − 2 a b ​ ) . If a > 0 a > 0 a > 0 , range is [ k , ∞ ) [k, \infty) [ k , ∞ ) ; if a < 0 a < 0 a < 0 , range is ( − ∞ , k ] (-\infty, k] ( − ∞ , k ] .

Absolute Value Functions ( f ( x ) = ∣ x ∣ + k f(x) = |x| + k f ( x ) = ∣ x ∣ + k ): Range is [ k , ∞ ) [k, \infty) [ k , ∞ ) .

  • 6. Function Evaluation and Composition
  • Evaluating Functions

Evaluating f ( a ) f(a) f ( a ) means replacing every instance of the variable x x x in f ( x ) f(x) f ( x ) with the input value or expression a a a .

Composite Functions

Given two functions f ( x ) f(x) f ( x ) and g ( x ) g(x) g ( x ) , the composite function ( f ∘ g ) ( x ) (f \circ g)(x) ( f ∘ g ) ( x ) , read as " f f f of g g g of x x x ", is defined by: ( f ∘ g ) ( x ) = f ( g ( x ) ) (f \circ g)(x) = f(g(x)) ( f ∘ g ) ( x ) = f ( g ( x ))

The output of the inner function g ( x ) g(x) g ( x ) becomes the direct input for the outer function f ( x ) f(x) f ( x ) .

Worked Step-by-Step Example

Let f ( x ) = x 2 − 3 x + 2 f(x) = x^2 - 3x + 2 f ( x ) = x 2 − 3 x + 2 and g ( x ) = 2 x + 1 g(x) = 2x + 1 g ( x ) = 2 x + 1 .

Evaluate g ( 3 ) g(3) g ( 3 ) : g ( 3 ) = 2 ( 3 ) + 1 = 7 g(3) = 2(3) + 1 = 7 g ( 3 ) = 2 ( 3 ) + 1 = 7

Evaluate ( f ∘ g ) ( 3 ) = f ( g ( 3 ) ) = f ( 7 ) (f \circ g)(3) = f(g(3)) = f(7) ( f ∘ g ) ( 3 ) = f ( g ( 3 )) = f ( 7 ) : f ( 7 ) = ( 7 ) 2 − 3 ( 7 ) + 2 = 49 − 21 + 2 = 30 f(7) = (7)^2 - 3(7) + 2 = 49 - 21 + 2 = 30 f ( 7 ) = ( 7 ) 2 − 3 ( 7 ) + 2 = 49 − 21 + 2 = 30

Find Algebraic Expression for ( f ∘ g ) ( x ) (f \circ g)(x) ( f ∘ g ) ( x ) : ( f ∘ g ) ( x ) = f ( 2 x + 1 ) = ( 2 x + 1 ) 2 − 3 ( 2 x + 1 ) + 2 (f \circ g)(x) = f(2x + 1) = (2x + 1)^2 - 3(2x + 1) + 2 ( f ∘ g ) ( x ) = f ( 2 x + 1 ) = ( 2 x + 1 ) 2 − 3 ( 2 x + 1 ) + 2 = ( 4 x 2 + 4 x + 1 ) − ( 6 x + 3 ) + 2 = 4 x 2 − 2 x = (4x^2 + 4x + 1) - (6x + 3) + 2 = 4x^2 - 2x = ( 4 x 2 + 4 x + 1 ) − ( 6 x + 3 ) + 2 = 4 x 2 − 2 x