Coordinate Geometry

Select a difficulty level to start practicing. Easy exams are free. Medium and hard exams require a premium subscription.

Easy examsMedium examsHard exams

Practice Coordinate Geometry with tests designed for the University Practice. Each question includes a full explanation so you can learn from every mistake. Mastering this Mathematics topic is key to improving your score on the entrance exam.

Coordinate Geometry accounts for about 15-20% of the ACT Math section. This area tests your ability to work with points, lines, curves, and shapes on the coordinate plane.

Distance and Midpoint Formulas

These two formulas are essential for coordinate geometry:
Distance between (x₁, y₁) and (x₂, y₂): d = √((x₂−x₁)² + (y₂−y₁)²)
Midpoint: ((x₁+x₂)/2, (y₁+y₂)/2)

Example 1: Find the distance and midpoint between A(1, 3) and B(7, 11).
Distance: √((7−1)² + (11−3)²) = √(36 + 64) = √100 = 10
Midpoint: ((1+7)/2, (3+11)/2) = (4, 7)

Example 2: The endpoints of a diameter are (−2, 5) and (6, 1). Find the center and radius of the circle.
Center = midpoint = ((−2+6)/2, (5+1)/2) = (2, 3)
Radius = half the diameter = √((6−(−2))² + (1−5)²)/2 = √(64+16)/2 = √80/2 = 4√5/2 = 2√5

Example 3: Point M(3, 7) is the midpoint of segment PQ. If P = (1, 4), find Q.
Using midpoint formula: (1 + x)/2 = 3 → x = 5 and (4 + y)/2 = 7 → y = 10. So Q = (5, 10).

Slope and Equations of Lines

Types of Slope Positive Rises left to right Negative Falls left to right Zero Horizontal line Undefined Vertical line Slope: m = (y₂ − y₁) / (x₂ − x₁) Slope-intercept: y = mx + b Point-slope: y − y₁ = m(x − x₁) Standard: Ax + By = C Parallel lines: m₁ = m₂ | Perpendicular lines: m₁ × m₂ = −1 Example 1: Write the equation of a line passing through (2, 5) with slope 3.
Using point-slope form: y − 5 = 3(x − 2) → y − 5 = 3x − 6 → y = 3x − 1

Example 2: Find the equation of a line perpendicular to y = 2x + 4 passing through (6, 1).
The original slope is 2, so the perpendicular slope is −1/2.
y − 1 = −1/2(x − 6) → y = −1/2 x + 3 + 1 → y = −1/2 x + 4

Example 3: Find the slope of the line 3x − 5y = 15.
Solve for y: −5y = −3x + 15 → y = (3/5)x − 3. Slope = 3/5.

Example 4: Line A passes through (0, 4) and (3, 10). Line B passes through (1, 2) and (4, 8). Are they parallel?
Slope of A = (10−4)/(3−0) = 6/3 = 2. Slope of B = (8−2)/(4−1) = 6/3 = 2. Same slope → yes, they are parallel.

Circles in Coordinate Geometry

Standard form: (x − h)² + (y − k)² = r², centered at (h, k) with radius r.

If the equation is in general form (x² + y² + Dx + Ey + F = 0), complete the square to find the center and radius.

Example 1: x² + y² − 6x + 4y − 12 = 0
Group and complete the square: (x² − 6x + 9) + (y² + 4y + 4) = 12 + 9 + 4
(x − 3)² + (y + 2)² = 25 → Center: (3, −2), Radius: 5

Example 2: Does the point (1, 6) lie inside, on, or outside the circle (x − 3)² + (y − 4)² = 9?
Substitute: (1−3)² + (6−4)² = 4 + 4 = 8. Since 8 < 9, the point lies inside the circle.

Circle: General Form to Standard Form General Form x² + y² + Dx + Ey + F = 0 Center and radius are hidden → Standard Form (x−h)² + (y−k)² = r² Center (h,k), Radius r How to Complete the Square 1. Group x-terms and y-terms. Move constant to right side. 2. For each group, add (half the coefficient)² to BOTH sides.

Parabolas

y = ax² + bx + c opens up (a > 0) or down (a < 0). Vertex at x = −b/(2a).
Vertex form: y = a(x − h)² + k with vertex (h, k).

Parabola: y = (x − 2)² − 3 Vertex (2, −3) x = 2 Key Features Vertex: (h, k) = (2, −3) Axis of symmetry: x = 2 Opens upward (a = 1 > 0) y-intercept: (0−2)² − 3 = 1 Min value = −3 Example: Find the vertex of y = −2x² + 12x − 7.
x = −12/(2(−2)) = −12/(−4) = 3. y = −2(9) + 12(3) − 7 = −18 + 36 − 7 = 11. Vertex: (3, 11). Since a = −2 < 0, the parabola opens down and the vertex is a maximum.

Transformations

  • Translation: shifts the graph without changing shape
  • Reflection over x-axis: (x, y) → (x, −y)
  • Reflection over y-axis: (x, y) → (−x, y)
  • f(x) + k shifts up k; f(x − h) shifts right h
Memory aid: Changes inside f(x) affect x and work opposite to intuition. Changes outside affect y and work as expected.

Example: The graph of y = f(x) is shifted right 3 units and up 2 units. What is the new equation?
y = f(x − 3) + 2. (Right 3 → subtract 3 inside; up 2 → add 2 outside.)

Linear Inequalities and Regions

Graph the boundary line, then shade the region that satisfies the inequality. Use a test point (often the origin) to determine which side to shade.

Example: Graph y > 2x − 1.
Draw the dashed line y = 2x − 1 (dashed because > is strict, not ≥). Test (0, 0): 0 > 2(0) − 1 → 0 > −1, true. Shade the side containing the origin.

Common Mistakes

Top 5 Coordinate Geometry Traps 1. Forgetting the negative in standard circle form: (x−3)² means center x = +3, not −3. 2. Mixing up slope of parallel (same) vs. perpendicular (negative reciprocal). 3. Transformation direction: f(x−3) shifts RIGHT 3 (opposite of the minus sign). 4. Computing distance as (x₂−x₁) + (y₂−y₁) instead of using the distance formula with squares. 5. Forgetting to add completed-square values to BOTH sides when converting circle equations.

Quick Reference — Formula Sheet

Coordinate Geometry Quick Reference Distance d = √((x₂−x₁)² + (y₂−y₁)²) Midpoint M = ((x₁+x₂)/2, (y₁+y₂)/2) Slope m = (y₂−y₁)/(x₂−x₁) Parallel: m₁ = m₂ Perpendicular: m₁·m₂ = −1 Line Forms y = mx + b (slope-intercept) y−y₁ = m(x−x₁) (point-slope) Ax + By = C (standard) Circle (x−h)² + (y−k)² = r² Center (h,k), Radius r Parabola Vertex form: y = a(x−h)² + k Vertex: x = −b/(2a) a>0 opens up, a<0 opens down Transformations f(x−h): right h units f(x)+k: up k units −f(x): reflect over x-axis f(−x): reflect over y-axis

ACT Strategies for Coordinate Geometry

  • Memorize the big three: Distance, midpoint, and slope formulas appear in nearly every coordinate geometry problem.
  • Parallel/Perpendicular: Find the slope first, then use point-slope form for the new equation.
  • Complete the square: When given a circle in general form, complete the square to identify center and radius.
  • Transformation rule: Inside changes → opposite direction. Outside changes → same direction.
  • Intersection of lines: Set the equations equal and solve for x, then find y.
  • Sketch it: Draw a quick diagram when possible — it helps catch errors and eliminate wrong answers.
  • Vertex formula: x = −b/(2a) is the fastest way to find a parabola's vertex on the ACT.
  • Rearrange to slope-intercept: If the line is in standard form, convert to y = mx + b to instantly read slope and y-intercept.

Practice Walkthrough

ACT-Style Problem: What is the equation of the line that is perpendicular to 4x + 2y = 10 and passes through the point (−2, 3)?

A. y = −2x − 1    B. y = 1/2 x + 4    C. y = 2x + 3    D. y = −1/2 x + 2

Step 1 — Find the original slope: 4x + 2y = 10 → 2y = −4x + 10 → y = −2x + 5. Slope = −2.
Step 2 — Find perpendicular slope: Negative reciprocal of −2 is 1/2.
Step 3 — Use point-slope form: y − 3 = 1/2(x − (−2)) → y − 3 = 1/2(x + 2) → y = 1/2 x + 1 + 3 → y = 1/2 x + 4.
Elimination: Choice A has slope −2 (parallel, not perpendicular). Choice C has slope 2 (wrong reciprocal — forgot the negative). Choice D has slope −1/2 (negative of the correct answer). Only B (y = 1/2 x + 4) is correct.
← PreviousMathematicsNext →Easy exams