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Bunuel
If a triangle in the xy-coordinate system has vertices at (-2 , -3), (4, -3) and (28, 7), what is the area of the triangle?

A. 30
B. 36
C. 48
D. 60
E. 65

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answer A
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Bunuel
If a triangle in the xy-coordinate system has vertices at (-2 , -3), (4, -3) and (28, 7), what is the area of the triangle?

A. 30
B. 36
C. 48
D. 60
E. 65

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(-2,-3) is in 3rd quadrant, (4,-3) is in 4th quadrant while (28,7) is in 1st quadrant.
Also, y coordinate of (-2,-3) and (4,-3) is same so they lie on same horizontal line.

So base of triangle = sqrt[(-3+3)^2 + (4+2)^2] = 6

And height of triangle = 7 - (-3)
= 10

So area of triangle = 1/2*6*10
= 30
Hence option (A),

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Bunuel
If a triangle in the xy-coordinate system has vertices at (-2 , -3), (4, -3) and (28, 7), what is the area of the triangle?

A. 30
B. 36
C. 48
D. 60
E. 65

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Solution:

We can let A = (-2 , -3), B = (4, -3), and C = (28, 7). Since A and B have the same y-coordinates, we should let AB be the base of the triangle and hence AB = |4 - (-2)| = 6. Now, no matter where C is, the height of the triangle is the absolute difference between the y-coordinates of A (or B) and C. That is, the height of the triangle is |7 - (-3)| = 10. Therefore, the area of the triangle is ½ x 6 x 10 = 30.

Answer: A
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