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The points A(0, 0), B(0, 4a - 5) and C(2a + 1, 2a + 6) form

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The points A(0, 0), B(0, 4a - 5) and C(2a + 1, 2a + 6) form [#permalink] New post 29 Oct 2017, 01:53
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Question Stats:

60% (00:55) correct 40% (01:23) wrong based on 5 sessions
The points A(0, 0), B(0, 4a - 5) and C(2a + 1, 2a + 6) form a triangle. If angle ABC = 90, what is the area of triangle ABC?

A. 102
B. 120
C. 132
D. 144
E. 156

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[Reveal] Spoiler: OA
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Re: The points A(0, 0), B(0, 4a - 5) and C(2a + 1, 2a + 6) form [#permalink] New post 30 Oct 2017, 09:04
Not easy to explain without a drawing but I'll try.

Point A (0,0) is the origin. Then, point B (0, 4a-5) is on the y axis, while point C (2a+1, 2a+6) must have the same y coordinate of B in order to angle ABC to be 90°.

Thus, we can retrieve the value of a equating 4a-5 = 2a+6, getting a = 11/2. Then, point B coordinates becomes (0, 17) and point C coordinates become (12, 17).

Then, the area of the triangle is computed as AB*BC/2 = 17*12/2 = 102.

Answer A
Re: The points A(0, 0), B(0, 4a - 5) and C(2a + 1, 2a + 6) form   [#permalink] 30 Oct 2017, 09:04
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The points A(0, 0), B(0, 4a - 5) and C(2a + 1, 2a + 6) form

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