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by jainrahul1985 » Thu Aug 18, 2011 9:01 am
The coordinates of a right triangle are (-2, 4), (-6, -6), (3,2), find the area of the triangle ?

A. √ 29 B. 29 C. √ 27 D. 27 E. 31

OA B
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by Anurag@Gurome » Thu Aug 18, 2011 9:12 am
jainrahul1985 wrote:The coordinates of a right triangle are (-2, 4), (-6, -6), (3,2), find the area of the triangle ?
There is a formula to calculate the area of a triangle ABC,
Image
Where, Ax is the x coordinate of A and so on...

Therefore, area of the triangle = |(-2)*(-6 - 2) + (-6)*(2 - 4) + 3*(4 - (-6))|/2 = |16 + 12 + 30|/2 = 58/2 = 29

The correct answer is B.

Also we can utilize the fact that the triangle is right angled one. In that case the area of triangle = (Product of the smaller two sides)/2
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by ashutoshkumar7 » Thu Aug 18, 2011 9:30 am
Just in case you do not remember the formula:
seeing the co-ordinates you can see two of the lines are perpendicular. as their slope is - reciprocal.

the line joining points (-2,4), (-6,-6) and line joining points (-2,4), (3,2)
knowing this all we have to do is find the distance of these two line and use the formula of 1/2*base*height to get the area.

Distance between any two points is Sqr Root((Y1 -Y2)sqr -(X1-X2)sqr)

So the length are root(116) and root(29)

so the area is root(116) *root(29) /2 = 29

I do not know how to put symbols for root and square :)
Thanks,
Ashutosh
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