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

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Solution

Let D be the mid- point of BC. Then, the coordinates of D are (4, 0).

We have,

∴ Area of $Δ$ ABC = $\frac{1}{2}$ |(4 × (-2) + 3 × 2 + 5 × (-6)) - (3 × (-6) + 5 × (-2) + 4 × 2)|

⇒ Area of $Δ$ ABC = $\frac{1}{2}$ |(-8 + 6 - 30) - (-18 - 10 + 8)|

⇒ Area of $Δ$ ABC = $\frac{1}{2}$ |-32 + 20| = 6 sq. units

Also, We have

∴ Area of $Δ$ ABD = $\frac{1}{2}$ |{(4 × (-2) + 3 × 0 + 4 ×(-6))} - {3 × (-6) + 4 × (-2) + 4 × 0}|

⇒ Area of $Δ$ ABD = $\frac{1}{2}$ |(-8 + 0 + 26) - (-18 - 8 + 0)|

⇒ Area of $Δ$ ABD = $\frac{1}{2}$ |-32 + 26| = 3 sq. units

⇒ $\frac{A r e a o f Δ A B C}{A r e a o f Δ A B D}$ = $\frac{6}{3}$ = $\frac{2}{1}$

⇒ Area of $Δ$ ABC = 2(Area of $Δ$ ABD)

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