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📑 In This Chapter Guide (Table of Contents)
1. The Distance Formula & Applications
The straight-line distance between two coordinates A(x₁, y₁) and B(x₂, y₂) in the Cartesian plane is derived from Pythagoras Theorem:
AB = √[ (x₂ - x₁)² + (y₂ - y₁)² ]
Distance from Origin (0,0): Distance of point P(x, y) from origin is simply OP = √(x² + y²).
Proving Geometric Figures on Cartesian Plane:
- Equilateral Triangle: Show all three side lengths are equal:
AB = BC = CA. - Right-Angled Triangle: Show sum of squares of two sides equals square of the third side:
AB² + BC² = AC². - Square vs. Rhombus: All 4 sides equal (
AB = BC = CD = DA). For a Square, diagonals are equal (AC = BD); for a Rhombus, diagonals are NOT equal. - Rectangle vs. Parallelogram: Opposite sides equal. For a Rectangle, diagonals are equal; for a Parallelogram, diagonals are unequal.
2. Section Formula & Midpoint Coordinates
The coordinates of point P(x, y) that divides the line segment joining A(x₁, y₁) and B(x₂, y₂) internally in the ratio m₁ : m₂ are given by:
x = (m₁x₂ + m₂x₁) / (m₁ + m₂)
y = (m₁y₂ + m₂y₁) / (m₁ + m₂)
Special Case 1: Midpoint Formula (Ratio 1 : 1):
Midpoint M = [ (x₁ + x₂) / 2, (y₁ + y₂) / 2 ]
Special Case 2: Centroid of a Triangle:
The centroid G of ΔABC with vertices (x₁, y₁), (x₂, y₂), (x₃, y₃) is:
G = [ (x₁ + x₂ + x₃) / 3, (y₁ + y₂ + y₃) / 3 ]
💡 Frequently Asked Questions (FAQ)
❓ Find the ratio in which the Y-axis divides the line segment joining A(5, -6) and B(-1, -4).
Any point on the Y-axis has x-coordinate equal to 0, i.e., P(0, y). Let the ratio be k : 1. Using section formula for x: 0 = [k(-1) + 1(5)] / (k + 1) ⟹ -k + 5 = 0 ⟹ k = 5. Therefore, the Y-axis divides the segment in the ratio 5 : 1.
❓ Show that points A(1, 5), B(2, 3), and C(-2, -11) are collinear.
Calculate distances: AB = √[(2-1)² + (3-5)²] = √[1 + 4] = √5. BC = √[(-2-2)² + (-11-3)²] = √[16 + 196] = √212 = 2√53. AC = √[(-2-1)² + (-11-5)²] = √[9 + 256] = √265. Since AB + BC ≠ AC, the points are NOT collinear.
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