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Question

The area of triangle with vertices A(x1,y1),B(x2,y2) and C(x3,y3) is:


A
12[x1(y2y3)+x2(y3y1)+x3(y1y2)]
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B

12[x1(y2y3)+x3(y3y1)+x1(y1y2)]

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C

12[x2(y2y3)+x2(y3y1)+x3(y1y2)]

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D
All of the above.
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Solution

The correct option is A 12[x1(y2y3)+x2(y3y1)+x3(y1y2)]


Let ABC be any triangle whose vertices are A(x1,y1),B(x2,y2)and C(x3,y3). Draw AP, BQ and CR perpendiculars from A, B and C, respectively, to the x-axis. Clearly ABQP, APRC and BQRC are all trapeziums.
Now from the figure, it is clear that
Area of ABC = Area of trapezium ABQP + Area of trapezium APRC - Area of trapezium BQRC.
Area of a trapezium =12 (Sum of parallel sides)(Distance between the parallel sides)
Area of ΔABC=12(BQ+AP)QP+12(AP+CR)PR12(BQ+CR)QR
=12(y2+y1)(x1x2)+12(y3+y1)(x3x1)12(y2+y3)(x3x2)
=12[x1(y2y3)+x2(y3y1)+x3(y1y2)]
Thus, the area of ΔABC is the numerical value of the expression 12[x1(y2y3)+x2(y3y1)+x3(y1y2)]


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