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Coordinate Geometry Lecture Summary

Feb 11, 2025

Lecture Notes: Coordinate Geometry Problems

Key Concepts

  • Coordinate Geometry involves solving problems related to points, lines, and shapes in a coordinate plane.

Problem 1: Coordinates of Point B

  • Vertical Line: Points on a vertical line share the same x-coordinate.
  • Horizontal Line: Points on a horizontal line share the same y-coordinate.
  • Solution: For point B, x = 1 (same as A), y = 2 (same as C).

Problem 2: Area of a Triangle

  • Formula: Area = 1/2 * base * height.
  • Base: Distance between B (1,2) and C (5,2) = 4 units.
  • Height: Distance between A (1,5) and B (1,2) = 3 units.
  • Area: 1/2 * 4 * 3 = 6 square units.

Problem 3: Circle Center and Properties

Part A: Center of Circle

  • Midpoint Formula: Midpoint = ((x1 + x2)/2, (y1 + y2)/2).
  • Center of circle (midpoint of diameter): (1+7)/2, (2+10)/2 = (4, 6).

Part B: Area & Circumference

  • Radius: Use distance formula.
    • Distance = sqrt((7-4)^2 + (10-6)^2) = 5.
  • Area: πr² = 25π.
  • Circumference: 2πr = 10π.

Part C: Standard Equation of Circle

  • Equation: (x-h)² + (y-k)² = r².
  • For center (4,6) and radius 5: (x-4)² + (y-6)² = 25.

Problem 4: Tangent Line to Circle

  • Tangent Line: Perpendicular to radius at point of tangency.
  • Slope of AB: (8-2)/(5-3) = 3.
  • Perpendicular Slope: Negative reciprocal = -1/3.
  • Equation of Tangent Line: Using point-slope form with point B (5,8).
    • Equation: y = -1/3x + 29/3.

Problem 5: Area Bounded by Axes and Line

  • Equation: 4x + y = 8.
  • Intercepts: x-intercept (2,0), y-intercept (0,8).
  • Triangle Area: 1/2 * base * height = 8 square units.

Problem 6: Plotting 3D Points and Distances

  • Point: (3,4,5).
  • Distance Formula: Between origin (0,0,0) and point (3,4,5).
    • Distance = sqrt(3² + 4² + 5²) = 5√2 ≈ 7.07.

Problem 7: Distance from a Point to a Line

  • Point: (5,3); Line: 3x + 4y - 7 = 0.
  • Distance Formula: |Ax1 + By1 + C| / sqrt(A² + B²).
    • Distance = 4 units.

Problem 8: Area of Shaded Region (Circle and Square)

  • Square Side: 8 units (diameter of circle).
  • Circle Radius: 4 units.
  • Area Calculation: Square area - Circle area = 64 - 16π ≈ 13.73.

Problem 9: Equilateral Triangle Area

Part A: Area Calculation

  • Side Length: 8 units.
  • Area Formula: (sqrt(3)/4) * s².
  • Area: 16√3 square units.*

Part B: Coordinates of Point B

  • Using 30-60-90 Triangle Properties:
    • Coordinates: (4, 4√3).

Problem 10: Triangle Medians and Bisectors

Part A: Median to Segment AC

  • Midpoint of AC: Calculated using midpoint formula.
  • Equation of Median: y = 7x - 25.

Part B: Perpendicular Bisector of AC

  • Slope of Perpendicular Line: Negative reciprocal of AC slope.
  • Equation: y = -3x + 15.

Part C: Altitude from B to AC

  • Shares Slope: With perpendicular bisector.
  • Equation: y = -3x + 25.

These notes summarize the lecture, organizing key concepts and solutions for each problem in coordinate geometry. They include explanations of methodologies such as using formulas for midpoints, slopes, and areas, which are essential to solving these problems.

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