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What is the area of the triangle ABC with A(e,p) B(2e,3p)
and C(3e,5p)?
where p = PI (3.141592654)
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Answer / guest
A tricky ONE.
Given 3 points are colinear. Hence, it is a straight line.
Hence area of triangle is 0.
| Is This Answer Correct ? | 10 Yes | 2 No |
By the co-ordination of A(e,p) B(2e,3p)
and C(3e,5p) are A(x1,y1),B(x2,y2) and C(x3,y3) respectively
then
Area of triangle=[(x1(y2-y3)+x2(y3-y1)+x3(y1-y2)/2]
=[(e(3p-5p)+2e(5p-p)+3e(p-3p)/2]
=0
Therefore in a given co ordinates area of triangle=0
| Is This Answer Correct ? | 5 Yes | 1 No |
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