1.6 Coordinate Geometry: Lines, Distance, Midpoints & Conic Sections

Ateneo College Entrance Test ACET

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1 Chapter 1: Mathematics

1.1 Real Numbers, Ratios, Rates & Percentages 1.2 Algebraic Expressions, Polynomials & Linear Equations 1.3 Functions, Domain/Range, Graphs & Transformations 1.4 Quadratic, Polynomial, Exponential & Logarithmic Equations 1.5 Plane Geometry: Triangles, Circles, Polygons & Perimeter/Area 1.6 Coordinate Geometry: Lines, Distance, Midpoints & Conic Sections 1.7 Trigonometry: Right Triangle Ratios, Unit Circle & Basic Identities 1.8 Arithmetic & Geometric Sequences, Series & Word Patterns 1.9 Permutations, Combinations, Probability & Statistics

2 Chapter 2: English Proficiency

2.1 Subject-Verb Agreement, Verb Tenses & Aspect 2.2 Pronoun-Antecedent Agreement, Case, Misplaced & Dangling Modifiers 2.3 Sentence Structure, Run-Ons, Fragments & Parallelism 2.4 Contextual Vocabulary, Word Roots, Prefixes & Suffixes 2.5 Advanced Synonyms, Antonyms & Nuanced Vocabulary 2.6 Verbal Analogies: Relationship Types & Solving Strategies 2.7 Punctuation, Capitalization, Mechanics & Common Errors 2.8 Diction, Idiomatic Expressions, Prepositions & Frequently Confused Words

3 Chapter 3: Reading Comprehension

3.1 Identifying Main Ideas, Central Themes & Paragraph Summaries 3.2 Locating Explicit Details, Supporting Evidence & Data 3.3 Drawing Logical Inferences & Unstated Conclusions 3.4 Analyzing Author's Purpose, Point of View, Tone & Attitude 3.5 Interpreting Figurative Language, Tone Words & Context Clues 3.6 Text Structure, Rhetorical Devices & Evaluating Arguments

4 Chapter 4: Logical and Abstract Reasoning

4.1 Number & Letter Series Patterns & Formulas 4.2 Abstract Reasoning: Matrix Patterns & Figure Rotations 4.3 Formal Syllogisms, Categorical Logic & Deductive Proofs 4.4 Coding-Decoding, Cipher Patterns & Symbol Operations 4.5 Spatial Visualization, Paper Folding & Venn Diagram Logic

5 Chapter 5: General Knowledge

5.1 Philippine History, Government, Constitution & Civics 5.2 Integrated Science: Biology, Chemistry, Physics & Earth Science 5.3 Philippine & World Geography, Basic Economics & Cultural Facts 5.4 Current Events, World Landmarks, Literature & General Trivia

6 Chapter 6: ACET Essay Writing & Exam Strategies

6.1 ACET Essay: Thesis Development, Structuring & Argumentation 6.2 Time Allocation, Pacing Strategies & Subtest Tactics 6.3 Guessing Strategies (No Penalty), Test Anxiety & Exam Day Readiness

1.6 Coordinate Geometry: Lines, Distance, Midpoints & Conic Sections

Key Takeaways Explain This Topic AI

Calculate distance, midpoint, and slope in the Cartesian plane using explicit coordinate formulas.

Derive line equations using slope-intercept, point-slope, and standard form, enforcing parallel/perpendicular slope relationships.

Calculate the perpendicular distance from a point to a line using the standard distance formula.

Identify conic sections (circles, parabolas, ellipses, hyperbolas) from general quadratic equations using the discriminant B^2 - 4AC.

Last updated: July 2026

Coordinate Geometry: Lines, Distance, Midpoints & Conic Sections

Coordinate geometry bridges algebra and plane geometry, representing one of the most heavily tested content domains on the Ateneo College Entrance Test (ACET) Mathematics subtest. Success on the ACET requires rapid execution of distance and midpoint formulas, linear slope derivations, perpendicular distance calculations from points to lines, Shoelace polygon area determinations, and structural identification of conic sections (circles, parabolas, ellipses, hyperbolas).

1. Fundamental Cartesian Formulas

Given two arbitrary points in the Cartesian plane 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 ​ ) :

Primary Metric Formulas

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 ​

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 ​ ​ )

Slope Formula: 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 ​ )

Section Formula (Internal Partitioning ratio m : n m:n m : n ): P = ( m x 2 + n x 1 m + n , m y 2 + n y 1 m + n ) P = \left( \frac{m x_2 + n x_1}{m + n}, \frac{m y_2 + n y_1}{m + n} \right) P = ( m + n m x 2 ​ + n x 1 ​ ​ , m + n m y 2 ​ + n y 1 ​ ​ )

Centroid of a Triangle ( x 1 , y 1 ) , ( x 2 , y 2 ) , ( x 3 , y 3 ) (x_1, y_1), (x_2, y_2), (x_3, y_3) ( x 1 ​ , y 1 ​ ) , ( x 2 ​ , y 2 ​ ) , ( x 3 ​ , y 3 ​ ) : C = ( x 1 + x 2 + x 3 3 , y 1 + y 2 + y 3 3 ) C = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right) C = ( 3 x 1 ​ + x 2 ​ + x 3 ​ ​ , 3 y 1 ​ + y 2 ​ + y 3 ​ ​ )

• 2. Linear Equations, Parallelism & Perpendicular Distance
• Linear Forms
• Slope-Intercept Form: y = m x + b y = mx + b y = m x + b ( m m m is slope, b b b is y-intercept)
• Point-Slope Form: y − y 1 = m ( x − x 1 ) y - y_1 = m(x - x_1) y − y 1 ​ = m ( x − x 1 ​ )

General Standard Form: A x + B y + C = 0 Ax + By + C = 0 A x + B y + C = 0 (where slope m = − A / B m = -A/B m = − A / B , y-intercept = − C / B -C/B − C / B )

Parallel & Perpendicular Line Relationships

Parallel Lines ( L 1 ∥ L 2 L_1 \parallel L_2 L 1 ​ ∥ L 2 ​ ): Slopes are equal ( m 1 = m 2 m_1 = m_2 m 1 ​ = m 2 ​ ). Lines never intersect unless identical.

Perpendicular Lines ( L 1 ⊥ L 2 L_1 \perp L_2 L 1 ​ ⊥ L 2 ​ ): Slopes are negative reciprocals ( m 1 ⋅ m 2 = − 1 ⟹ m 2 = − 1 m 1 m_1 \cdot m_2 = -1 \implies m_2 = -\frac{1}{m_1} m 1 ​ ⋅ m 2 ​ = − 1 ⟹ m 2 ​ = − m 1 ​ 1 ​ ).

Perpendicular Distance from Point ( x 0 , y 0 ) (x_0, y_0) ( x 0 ​ , y 0 ​ ) to Line A x + B y + C = 0 Ax + By + C = 0 A x + B y + C = 0

The shortest distance from an external point ( x 0 , y 0 ) (x_0, y_0) ( x 0 ​ , y 0 ​ ) to a straight line expressed in general form A x + B y + C = 0 Ax + By + C = 0 A x + B y + C = 0 is: d = ∣ A x 0 + B y 0 + C ∣ A 2 + B 2 d = \frac{|A x_0 + B y_0 + C|}{\sqrt{A^2 + B^2}} d = A 2 + B 2 ​ ∣ A x 0 ​ + B y 0 ​ + C ∣ ​

Shoelace Formula for Polygon Area

To calculate the area of a polygon with ordered vertices ( x 1 , y 1 ) , ( x 2 , y 2 ) , … , ( x n , y n ) (x_1, y_1), (x_2, y_2), \dots, (x_n, y_n) ( x 1 ​ , y 1 ​ ) , ( x 2 ​ , y 2 ​ ) , … , ( x n ​ , y n ​ ) : Area = 1 2 ∣ ( x 1 y 2 + x 2 y 3 + ⋯ + x n y 1 ) − ( y 1 x 2 + y 2 x 3 + ⋯ + y n x 1 ) ∣ \text{Area} = \frac{1}{2} \left| (x_1 y_2 + x_2 y_3 + \dots + x_n y_1) - (y_1 x_2 + y_2 x_3 + \dots + y_n x_1) \right| Area = 2 1 ​ ∣ ( x 1 ​ y 2 ​ + x 2 ​ y 3 ​ + ⋯ + x n ​ y 1 ​ ) − ( y 1 ​ x 2 ​ + y 2 ​ x 3 ​ + ⋯ + y n ​ x 1 ​ ) ∣

3. Conic Sections: Canonical Equations & Structural Properties

A conic section is formed by the intersection of a plane and a double-napped cone. In general form without xy-rotation ( B = 0 B=0 B = 0 ), conics follow: A x 2 + C y 2 + D x + E y + F = 0 A x^2 + C y^2 + D x + E y + F = 0 A x 2 + C y 2 + D x + E y + F = 0

Conic Section Taxonomy & Properties Table

Conic Section Standard Canonical Equation Center / Vertex Key Metric Identities & Foci

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 ​ , Constant distance from center

Parabola (Vertical) ( x − h ) 2 = 4 p ( y − k ) (x-h)^2 = 4p(y-k) ( x − h ) 2 = 4 p ( y − k ) Vertex ( h , k ) (h,k) ( h , k ) Focus ( h , k + p ) (h, k+p) ( h , k + p ) , Directrix y = k − p y = k-p y = k − p , p ≠ 0 p \neq 0 p  = 0

Parabola (Horizontal) ( y − k ) 2 = 4 p ( x − h ) (y-k)^2 = 4p(x-h) ( y − k ) 2 = 4 p ( x − h ) Vertex ( h , k ) (h,k) ( h , k ) Focus ( h + p , k ) (h+p, k) ( h + p , k ) , Directrix x = h − p x = h-p x = h − p , p ≠ 0 p \neq 0 p  = 0

Ellipse (Horizontal) ( x − h ) 2 a 2 + ( y − k ) 2 b 2 = 1 \frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1 a 2 ( x − h ) 2 ​ + b 2 ( y − k ) 2 ​ = 1 Center ( h , k ) (h,k) ( h , k ) c 2 = a 2 − b 2 ( a > b ) c^2 = a^2 - b^2 \quad (a > b) c 2 = a 2 − b 2 ( a > b ) , Foci ( h ± c , k ) (h \pm c, k) ( h ± c , k )

Hyperbola (Horizontal) ( x − h ) 2 a 2 − ( y − k ) 2 b 2 = 1 \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 a 2 ( x − h ) 2 ​ − b 2 ( y − k ) 2 ​ = 1 Center ( h , k ) (h,k) ( h , k ) c 2 = a 2 + b 2 c^2 = a^2 + b^2 c 2 = a 2 + b 2 , Asymptotes y − k = ± b a ( x − h ) y-k = \pm \frac{b}{a}(x-h) y − k = ± a b ​ ( x − h )

Identifying Conics via Coefficients ( A x 2 + C y 2 + D x + E y + F = 0 A x^2 + C y^2 + D x + E y + F = 0 A x 2 + C y 2 + D x + E y + F = 0 )

Parabola: A = 0 A = 0 A = 0 OR C = 0 C = 0 C = 0 (contains exactly one squared term).

Circle: A = C ≠ 0 A = C \neq 0 A = C  = 0 (squared terms have identical non-zero coefficients with the same sign).

Ellipse: A ≠ C A \neq C A  = C , but A ⋅ C > 0 A \cdot C > 0 A ⋅ C > 0 (squared terms have different positive coefficients of the same sign).

Hyperbola: A ⋅ C < 0 A \cdot C < 0 A ⋅ C < 0 (squared terms have coefficients of opposite signs).

• Worked Step-by-Step ACET Exam Problems
• Problem 1: Perpendicular Distance from Point to Line

Find the exact perpendicular distance from point P ( 3 , − 2 ) P(3, -2) P ( 3 , − 2 ) to the line 5 x − 12 y + 7 = 0 5x - 12y + 7 = 0 5 x − 12 y + 7 = 0 .

Step 1: Extract general line coefficients and point coordinates. A = 5 A = 5 A = 5 , B = − 12 B = -12 B = − 12 , C = 7 C = 7 C = 7 , x 0 = 3 x_0 = 3 x 0 ​ = 3 , y 0 = − 2 y_0 = -2 y 0 ​ = − 2 .

Step 2: Substitute into point-to-line distance formula. d = ∣ A x 0 + B y 0 + C ∣ A 2 + B 2 = ∣ 5 ( 3 ) + ( − 12 ) ( − 2 ) + 7 ∣ 5 2 + ( − 12 ) 2 d = \frac{|A x_0 + B y_0 + C|}{\sqrt{A^2 + B^2}} = \frac{|5(3) + (-12)(-2) + 7|}{\sqrt{5^2 + (-12)^2}} d = A 2 + B 2 ​ ∣ A x 0 ​ + B y 0 ​ + C ∣ ​ = 5 2 + ( − 12 ) 2 ​ ∣5 ( 3 ) + ( − 12 ) ( − 2 ) + 7∣ ​

Step 3: Evaluate absolute numerator and radical denominator.

• Numerator: ∣ 15 + 24 + 7 ∣ = ∣ 46 ∣ = 46 |15 + 24 + 7| = |46| = 46 ∣15 + 24 + 7∣ = ∣46∣ = 46
• Denominator: 25 + 144 = 169 = 13 \sqrt{25 + 144} = \sqrt{169} = 13 25 + 144 ​ = 169 ​ = 13

Final Distance: d = 46 13 ≈ 3.538 d = \frac{46}{13} \approx 3.538 d = 13 46 ​ ≈ 3.538 .

Problem 2: Completing the Square to Analyze a Circle Equation

Determine the center ( h , k ) (h, k) ( h , k ) and radius r r r of the circle defined by x 2 + y 2 − 6 x + 8 y − 11 = 0 x^2 + y^2 - 6x + 8y - 11 = 0 x 2 + y 2 − 6 x + 8 y − 11 = 0 .

Step 1: Group x and y variables and isolate constant term. ( x 2 − 6 x ) + ( y 2 + 8 y ) = 11 (x^2 - 6x) + (y^2 + 8y) = 11 ( x 2 − 6 x ) + ( y 2 + 8 y ) = 11 .

Step 2: Complete the square for both variable groups.

For x x x : ( − 6 2 ) 2 = ( − 3 ) 2 = 9 \left(\frac{-6}{2}\right)^2 = (-3)^2 = 9 ( 2 − 6 ​ ) 2 = ( − 3 ) 2 = 9 .

For y y y : ( 8 2 ) 2 = ( 4 ) 2 = 16 \left(\frac{8}{2}\right)^2 = (4)^2 = 16 ( 2 8 ​ ) 2 = ( 4 ) 2 = 16 .

Add 9 + 16 = 25 9 + 16 = 25 9 + 16 = 25 to both sides of the equation: ( x 2 − 6 x + 9 ) + ( y 2 + 8 y + 16 ) = 11 + 9 + 16 (x^2 - 6x + 9) + (y^2 + 8y + 16) = 11 + 9 + 16 ( x 2 − 6 x + 9 ) + ( y 2 + 8 y + 16 ) = 11 + 9 + 16 .

Step 3: Factor squared terms into standard canonical form. ( x − 3 ) 2 + ( y + 4 ) 2 = 36 (x - 3)^2 + (y + 4)^2 = 36 ( x − 3 ) 2 + ( y + 4 ) 2 = 36 .

Step 4: Read center coordinates and radius.

Center ( h , k ) = ( 3 , − 4 ) (h, k) = (3, -4) ( h , k ) = ( 3 , − 4 ) .

Radius r = 36 = 6 r = \sqrt{36} = 6 r = 36 ​ = 6 .

ACET Speed Tactics & Traps Summary

Problem Type Common Student Trap ACET Speed Tactic

Perpendicular Lines Using − m -m − m instead of negative reciprocal − 1 / m -1/m − 1/ m Negative reciprocal flips numerator/denominator and sign

Circle Radius Forgetting to add completed terms to right side Add ( D / 2 A ) 2 + ( E / 2 C ) 2 (D/2A)^2 + (E/2C)^2 ( D /2 A ) 2 + ( E /2 C ) 2 to constant side immediately

Ellipse Foci Metric Confusing c 2 = a 2 − b 2 c^2 = a^2 - b^2 c 2 = a 2 − b 2 with hyperbola c 2 = a 2 + b 2 c^2 = a^2 + b^2 c 2 = a 2 + b 2 Ellipse subtracts ( a 2 − b 2 a^2 - b^2 a 2 − b 2 ); Hyperbola adds ( a 2 + b 2 a^2 + b^2 a 2 + b 2 )

Midpoint Evaluation Subtracting coordinates instead of taking average Midpoint is coordinate average: add values and divide by 2

Slope of Vertical Line Setting vertical slope m = 0 m = 0 m = 0 instead of undefined Horizontal line has m = 0 m=0 m = 0 ; Vertical line has undefined slope