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# The points A( 0, 0), B( 0, 4a – 5), and C( 2a + 1, 2a + 6) f

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The points A( 0, 0), B( 0, 4a – 5), and C( 2a + 1, 2a + 6) f [#permalink]  30 Nov 2018, 18:11
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87% (03:04) correct 12% (03:30) wrong based on 8 sessions
The points A( 0, 0), B( 0, 4a – 5), and C( 2a + 1, 2a + 6) form a triangle. If ∠ ABC = 90 °, what is the area of triangle ABC?

A. 102

B. 120

C. 132

D. 144

E. 156
[Reveal] Spoiler: OA

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Re: The points A( 0, 0), B( 0, 4a – 5), and C( 2a + 1, 2a + 6) f [#permalink]  06 Dec 2018, 01:37
As angle ABC is 90. This is a right triangle with base AB and height BC. So area of Triangle is 1/2 (AB). BC
BC is perpendicular to AB. And A is at origin (0,0).
The Y-Coordinates of B & C should be same as they are on the same line i.e BC.
4a-5 = 2a+6
a=11/2
So
A= (0,0)
B=(0,4a-5)=(0,17)
C=(2a+1,2a+6)=(12,17)
Find distances AB & BC
AB = 17
BC = 12
Area of triangle = 1/2 (17)(12) = 102
It's A.
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Re: The points A( 0, 0), B( 0, 4a – 5), and C( 2a + 1, 2a + 6) f [#permalink]  06 Dec 2018, 18:32
Expert's post
Carcass wrote:
The points A( 0, 0), B( 0, 4a – 5), and C( 2a + 1, 2a + 6) form a triangle. If ∠ ABC = 90 °, what is the area of triangle ABC?

A. 102

B. 120

C. 132

D. 144

E. 156

90° is at B so area will be (AB*BC)/2
Since coordinates of x in A and B are the same that is 0, meaning A and B lie on y-axis, and 90° is at B, the y-coordinates of B and C should be the same.
So $$4a-5=2a+6.......2a=11.....a=\frac{11}{2}$$..
Now AB is nothing but y-coordinates of B and BC is nothing but x-coordinates of C
1)So X coordinates of C are $$2a+1=2*\frac{11}{2}+1=11+1=12$$
2) y coordinates of B = $$2a+6=2*\frac{11}{2}+6=11+6=17$$

Area = $$\frac{12*17}{2}=6*17=102$$

A
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Re: The points A( 0, 0), B( 0, 4a – 5), and C( 2a + 1, 2a + 6) f   [#permalink] 06 Dec 2018, 18:32
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