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M16-28

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Math Expert
Joined: 02 Sep 2009
Posts: 44351

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16 Sep 2014, 00:59
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Question Stats:

85% (01:38) correct 15% (01:10) wrong based on 68 sessions

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What is the area of a triangle enclosed by line $$2x+3y=6$$, line $$y-x=2$$ and $$X$$-axis on the coordinate plane?

A. 3
B. 4
C. 5
D. 8
E. 10
[Reveal] Spoiler: OA

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Math Expert
Joined: 02 Sep 2009
Posts: 44351

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16 Sep 2014, 00:59
Official Solution:

What is the area of a triangle enclosed by line $$2x+3y=6$$, line $$y-x=2$$ and $$X$$-axis on the coordinate plane?

A. 3
B. 4
C. 5
D. 8
E. 10

Look at the diagram below:

Lines $$y=-\frac{2}{3}x+2$$ and $$y=x+2$$ intersect at point $$(0, 2)$$. So the height of enclosed triangle is 2. Next, $$X$$-intercept of line $$y=-\frac{2}{3}x+2$$ is $$(3, 0)$$ and $$X$$-intercept of line $$y=x+2$$ is $$(-2, 0)$$, so the base of enclosed triangle is $$3-(-2)=5$$. The area is $$\frac{1}{2}*base*height=\frac{1}{2}*5*2=5$$.

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Posts: 405
Location: India
Concentration: Operations, Finance
GMAT 1: 670 Q48 V34
GPA: 3.8
WE: Operations (Commercial Banking)

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27 Aug 2016, 05:09
I think this is a high-quality question and I agree with explanation.
Manager
Joined: 23 Nov 2016
Posts: 76
Location: United States (MN)
GMAT 1: 760 Q50 V42
GPA: 3.51

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26 Feb 2017, 15:59
Bunuel wrote:
Official Solution:

What is the area of a triangle enclosed by line $$2x+3y=6$$, line $$y-x=2$$ and $$X$$-axis on the coordinate plane?

A. 3
B. 4
C. 5
D. 8
E. 10

Look at the diagram below:

Lines $$y=-\frac{2}{3}x+2$$ and $$y=x+2$$ intersect at point $$(0, 2)$$. So the height of enclosed triangle is 2. Next, $$X$$-intercept of line $$y=-\frac{2}{3}x+2$$ is $$(3, 0)$$ and $$X$$-intercept of line $$y=x+2$$ is $$(-2, 0)$$, so the base of enclosed triangle is $$3-(-2)=5$$. The area is $$\frac{1}{2}*base*height=\frac{1}{2}*5*2=5$$.

If I were to set up a matrix with the three vertices of the triangle (-2,0), (0,2), (3,0), and divide the determinant of that matrix by 2, that would give the area of the triangle. However, I also believe that I could simply set up a 3x3 matrix with the equation of each line, and use that matrix to also find the area by dividing the absolute value of its determinant by 2. The latter is faster for this problem because it saves me from finding each vertex.

| 2 3 6 |
|-1 1 2 |
| 0 1 0 |

This gives a determinant of -10, and half of the absolute value of that is 5. Is using equations for each line actually a viable way to get the area, or is the vertex method the only guaranteed method? It works for this problem but I haven't been able to find it actually documented as a common method. My apologies if I missed this somewhere.
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Concentration: Entrepreneurship, Technology
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21 Nov 2017, 10:36
Great Question! Beautiful
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Re: M16-28   [#permalink] 21 Nov 2017, 10:36
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