If A (4, –6), B (3, –2) and C (5, 2) are the vertices of triangle ABC, then verify the fact that a median of a triangle ABC divides it into two triangle of equal areas
Let D be the mid- point of BC. Then, the coordinates of D are (4, 0).
We have,
∴ Area of ΔABC = 12 |(4 × (-2) + 3 × 2 + 5 × (-6)) - (3 × (-6) + 5 × (-2) + 4 × 2)|
⇒ Area of ΔABC = 12 |(-8 + 6 - 30) - (-18 - 10 + 8)|
⇒ Area of ΔABC = 12|-32 + 20| = 6 sq. units
Also, We have
∴ Area of ΔABD = 12 |{(4 × (-2) + 3 × 0 + 4 ×(-6))} - {3 × (-6) + 4 × (-2) + 4 × 0}|
⇒ Area of ΔABD = 12 |(-8 + 0 + 26) - (-18 - 8 + 0)|
⇒ Area of ΔABD = 12|-32 + 26| = 3 sq. units
⇒ Area of ΔABCArea of ΔABD = 63 = 21
⇒ Area of ΔABC = 2(Area of ΔABD)