Introduction
Coordinate geometry connects algebra and geometry by placing points on the x-y plane. Skills like finding the distance between points, the slope of a line, and the area of a polygon (triangle) using coordinates are frequently tested in aptitude exams.
This pattern helps convert geometric questions into simple algebraic calculations using coordinate formulas.
Pattern: Coordinate Geometry (Distance, Slope, Area)
Pattern
Key concept: Use coordinate formulas to compute distances, slopes, midpoints and polygon areas directly from point coordinates.
Important formulas:
• Distance between (x₁, y₁) and (x₂, y₂): √[(x₂ - x₁)² + (y₂ - y₁)²].
• Slope of line through (x₁, y₁) and (x₂, y₂): (y₂ - y₁) / (x₂ - x₁) (vertical line → undefined).
• Midpoint of segment joining (x₁, y₁) and (x₂, y₂): ((x₁ + x₂)/2, (y₁ + y₂)/2).
• Area of triangle with vertices (x₁,y₁), (x₂,y₂), (x₃,y₃): ½ | x₁(y₂ - y₃) + x₂(y₃ - y₁) + x₃(y₁ - y₂) |.
Step-by-Step Example
Question
Given points A(-1, 2), B(3, -2) and C(5, 4):
(a) Find the distance AB.
(b) Find the slope of BC.
(c) Find the area of triangle ABC.
Solution
-
Step 1: Distance AB using distance formula.
AB = √[(x₂ - x₁)² + (y₂ - y₁)²] with A(-1,2), B(3,-2).
Compute differences: x₂ - x₁ = 3 - (-1) = 4; y₂ - y₁ = -2 - 2 = -4.
AB = √(4² + (-4)²) = √(16 + 16) = √32 = 4√2. -
Step 2: Slope of BC using slope formula.
B(3,-2), C(5,4). Slope m = (y₂ - y₁)/(x₂ - x₁) = (4 - (-2)) / (5 - 3) = 6/2 = 3. -
Step 3: Area of triangle ABC using determinant formula.
Area = ½ | x₁(y₂ - y₃) + x₂(y₃ - y₁) + x₃(y₁ - y₂) |.
Substitute A(-1,2), B(3,-2), C(5,4):
= ½ | (-1)((-2) - 4) + 3(4 - 2) + 5(2 - (-2)) |.
Simplify inside: (-1)(-6) + 3(2) + 5(4) = 6 + 6 + 20 = 32.
Area = ½ × |32| = 16 square units. -
Final Answers:
(a) AB = 4√2.
(b) Slope of BC = 3.
(c) Area of ΔABC = 16. -
Quick Check:
• Distance differences were equal magnitude → AB diagonal of a 4×4 right triangle → 4√2 makes sense.
• Slope 3 is integer and positive as C is above B and to the right.
• Area integer 16 is consistent with determinant arithmetic above ✅
Quick Variations
1. Find equation of line through two points (use point-slope form).
2. Check perpendicularity: slopes m₁·m₂ = -1.
3. Find midpoint and use as center for circles or reflections.
4. Area of polygon with >3 vertices: use shoelace formula (extension of triangle determinant).
Trick to Always Use
- Step 1 → Always compute coordinate differences first (x₂ - x₁, y₂ - y₁) to avoid sign mistakes.
- Step 2 → For slopes, simplify fraction early; watch for vertical (undefined) or horizontal (0) cases.
- Step 3 → For area, use the determinant form (or shoelace) and keep track of sign; take absolute value at the end.
Summary
Summary
Coordinate geometry problems reduce to straightforward algebra if you remember key formulas:
- Distance: √[(x₂ - x₁)² + (y₂ - y₁)²].
- Slope: (y₂ - y₁)/(x₂ - x₁) (vertical → undefined).
- Area of triangle: ½ | x₁(y₂ - y₃) + x₂(y₃ - y₁) + x₃(y₁ - y₂) |.
- Compute differences first, simplify fractions early, and always do a quick sanity check.
Hint: Subtract coordinates, square differences, add, then take square root and round neatly.
Common Mistakes: Leaving root form when numeric value is expected or rounding incorrectly.