Learn the Class 9 Madhyamik Mathematics Area of Triangle Region Formulas with step-by-step solved examples. This page covers the Area of a Triangle Formula, Collinearity Condition, and Centroid Formula using coordinates for quick revision and exam preparation.
Chapter 20 – Area of Triangle Region | Class 9 Flash Formula
Area of a Triangle
Area (A) = 1\over2 {x₁(y₂ − y₃) + x₂(y₃ − y₁) + x₃(y₁ − y₂)}
Example :
Find the area of the triangle whose vertices are A(1, 2), B(4, 6) and C(7, 2).
Solution
Area = 1\over2 × |1(6 − 2) + 4(2 − 2) + 7(2 − 6)|
= 1\over2 × |4 + 0 − 28|
= 1\over2 × 24 = 12 square units
Condition for Collinearity
Three points are collinear if
Area = 1\over2 × {x₁(y₂ − y₃) + x₂(y₃ − y₁) + x₃(y₁ − y₂)} = 0
Example :
Check whether A(1, 2), B(3, 4) and C(5, 6) are collinear.
Solution
Area = 1\over2 × {1(4 − 6) + 3(6 − 2) + 5(2 − 4)}
= 1\over2 × (−2 + 12 − 10)
= 1\over2 × 0
Therefore A, B and C are collinear.
Centroid of a Triangle
Centroid (G) = (x₁ + x₂ + x₃)\over3 , (y₁ + y₂ + y₃)\over3
Example :
Find the centroid of the triangle whose vertices are A(1, 2), B(4, 8) and C(7, 5).
Solution
x-coordinate = (1 + 4 + 7)\over3 = 12\over3 = 4
y-coordinate = (2 + 8 + 5)\over3 = 15\over3 = 5
Therefore, Centroid = (4, 5)
You have reached the end.