Question 139
If G is the centroid of the triangle $$\triangle ABC$$ with vertices A(-2, 3), B(-7. 5) and C(3, -5) then the area of $$\triangle GAB$$ (in sq. units) is
Solution
A = (-2, 3), B = (-7, 5) and C = (3, -5)
G = $$\frac{(-2) + (-7) + 3}{3}$$,$$\frac{3 + 5 + (-5)}{3}$$ = (-2, 1)
Area of $$\triangle GAB$$ = $$\frac{1}{2}$$[(-2)(5 - 1) + (-7)(1 - 3) + (-2)(3 - 5)]
= $$\frac{1}{2}$$[-8 + 14 + 4]
= $$\frac{1}{2}$$ * 10
= 5 sq. units
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