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Question

Find the area of a triangle whose vertices are (3,8), (-4,2) and (5,-1).

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Solution

Let A(3,8), B(-4,2) and C(5,-1) be the vertices of the given ABC. Then,
(x1=3,y1=8),(x2=4,y2=2),(x3=5,y3=1)
Area of ABC = 12|x1(y2y3)+x2(y3y1)+x3(y1y2)|

=12|3(3)4(9)+5(6)|

=12|9+36+30|=752=37.5 sq.units

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