find the area of the triangle whose vertices are (0,0), (a,0) and (0,b)
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3
Answer:
Area of triangle = 1/2[x1(y2-y3)+x2[y3-y1]+x3(y1-y2)]
A(0,0) B(a,0) and C(0,b)
=1/2[0(0-b)+a(b-0)+0(0-0)
=1/2[0+ab+0]
=1/2(ab)
=ab/2
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Answered by
1
Answer:
points are collinear....area is 0
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