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M16-28

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Bunuel
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M16-28 [#permalink] New post   16 Sep 2014, 00:59
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What is the area of the triangle formed by the lines \(2x + 3y = 6\), \(y - x = 2\), and the \(x\)-axis in the coordinate plane?

A. 3
B. 4
C. 5
D. 8
E. 10
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C
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Bunuel
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Re M16-28 [#permalink] New post   16 Sep 2014, 00:59
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Official Solution:

What is the area of the triangle formed by the lines \(2x + 3y = 6\), \(y - x = 2\), and the \(x\)-axis in the coordinate plane?

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


Refer to the diagram below:



The lines \(y = -\frac{2}{3}x + 2\) and \(y = x + 2\) intersect at point \((0, 2)\). This gives us the height of the enclosed triangle, which is 2 units. The \(x\)-intercepts of the lines are as follows: for the line \(y = -\frac{2}{3}x + 2\), it is \((3, 0)\), and for the line \(y = x + 2\), it is \((-2, 0)\). The base of the enclosed triangle is the distance between these two points, which is \(3 - (-2) = 5\) units. The area of the triangle can be calculated as \(\frac{1}{2}* \text{base} *\text{height} = \frac{1}{2}*5 *2 = 5\) square units.


Answer: C
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Re: M16-28 [#permalink] New post   27 Aug 2016, 05:09
I think this is a high-quality question and I agree with explanation.
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Re: M16-28 [#permalink] New post   26 Feb 2017, 15:59
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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 \(\frac{1}{2}*base*height=\frac{1}{2}*5*2=5\).


Answer: C


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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Re: M16-28 [#permalink] New post   21 Nov 2017, 10:36
Great Question! Beautiful
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Re: M16-28 [#permalink] New post   20 Jul 2019, 04:04
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 \(\frac{1}{2}*base*height=\frac{1}{2}*5*2=5\).


Answer: C



Hi, Did not understand the highlighted part. Derived the height but didn't get the base calc. Please help! How is X coordinate 3 and -2?

TIA
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Bunuel
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Re: M16-28 [#permalink] New post   20 Jul 2019, 04:13
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archimaitreya25 wrote:
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 \(\frac{1}{2}*base*height=\frac{1}{2}*5*2=5\).


Answer: C



Hi, Did not understand the highlighted part. Derived the height but didn't get the base calc. Please help! How is X coordinate 3 and -2?

TIA


The x-intercept of a line, (x, 0), is a point at which the line crosses the x-axis.

To get x-intercept of line \(2x+3y=6\), put y = 0 --> x = 3. So, the x-intercept of \(2x+3y=6\) is (3, 0).
To get x-intercept of line \(y-x=2\), put y = 0 --> x = -3. So, the x-intercept of \(y-x=2\) is (-2, 0).

The distance between, (3, 0) ans (-2, 0) = is the length of a base = 5.
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Re: M16-28 [#permalink] New post   20 Jul 2019, 05:27
Bunuel wrote:
archimaitreya25 wrote:
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 \(\frac{1}{2}*base*height=\frac{1}{2}*5*2=5\).


Answer: C



Hi, Did not understand the highlighted part. Derived the height but didn't get the base calc. Please help! How is X coordinate 3 and -2?

TIA


The x-intercept of a line, (x, 0), is a point at which the line crosses the x-axis.

To get x-intercept of line \(2x+3y=6\), put y = 0 --> x = 3. So, the x-intercept of \(2x+3y=6\) is (3, 0).
To get x-intercept of line \(y-x=2\), put y = 0 --> x = -3. So, the x-intercept of \(y-x=2\) is (-2, 0).

The distance between, (3, 0) ans (-2, 0) = is the length of a base = 5.



Thanks! Got it!
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Bunuel
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Re: M16-28 [#permalink] New post   26 Jan 2022, 11:54
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Similar questions to practice:



Hope it helps.
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Bunuel
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Re: M16-28 [#permalink] New post   28 Apr 2023, 13:31
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I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.
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Re: M16-28 [#permalink]
28 Apr 2023, 13:31
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