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In the coordinate plane above, if the equation of ℓ1 is y = x − 3 and

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In the coordinate plane above, if the equation of ℓ1 is y = x − 3 and [#permalink]

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In the coordinate plane above, if the equation of ℓ1 is y = x − 3 and the equation of ℓ2 is x = 2, then what is the area of the shaded region?

(A) 3
(B) 4
(C) \(3 \sqrt{2}\)
(D) 4.5
(E) 5

[Reveal] Spoiler:
Attachment:
2016-01-04_0026.png
2016-01-04_0026.png [ 9.48 KiB | Viewed 1910 times ]
[Reveal] Spoiler: OA

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Re: In the coordinate plane above, if the equation of ℓ1 is y = x − 3 and [#permalink]

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New post 04 Jan 2016, 00:45
I see that the shaded region is composed of two parts: a top rectangle and a bottom triangle. The final area will be the sum of those two parts.

Top rectangle:
The left bound is x = 0, right bound x = 2.
The top bound is y = 0, with the bottom bound at where our lines intersect. We calculate the intersection to be y = -1.
X = [0, 2]
Y = [0, -1]
Area for the rectangle: absolute(2 * -1) = 2.

Bottom triangle:
Our length is the same as the rectangle from above, 2.
To find our triangle height, we need to find where line 1 intersects with the origin. y = x - 3 -> y = -3.
Our triangle height is 3 - 1 = 2.
Our triangle area = (1/2) height*length -> (1/2) 2 * 2.
Area for the triangle = 2.

As I mentioned first in this post, our final area will be the area of the triangle + the area of the rectangle. Plugging in the calculated values from above, 2 + 2 = 4.

The answer should be B, 4.
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Re: In the coordinate plane above, if the equation of ℓ1 is y = x − 3 and [#permalink]

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New post 04 Jan 2016, 05:37
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Bunuel wrote:
Image
In the coordinate plane above, if the equation of ℓ1 is y = x − 3 and the equation of ℓ2 is x = 2, then what is the area of the shaded region?

(A) 3
(B) 4
(C) \(3 \sqrt{2}\)
(D) 4.5
(E) 5

[Reveal] Spoiler:
Attachment:
2016-01-04_0026.png


I'll calculate this one using the area formula of the trapezoid.

2: x=2 means the hieght of the trapezoid is equal to 2
1: y=x-3, to calculate the base we can set x=0 so y=-3 (Base 1)
To calculate the 2nd base we must find the point of intersection of lines ℓ1&2 --> y=2-3=-1 (Base 2)
We have everything we need to calculate the area:\(\frac{(Base1+ Base2)*Height}{2}\) = \(\frac{(3+1)*2}{2}=4\)
Answer B
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Re: In the coordinate plane above, if the equation of ℓ1 is y = x − 3 and [#permalink]

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New post 07 Dec 2017, 12:11
When we reverse the diagram for simplicity, the triangles formed have the following co-ordinates

Attachment:
Inverted diagram.png
Inverted diagram.png [ 2.67 KiB | Viewed 619 times ]


The shaded region's area is difference in the areas of the larger triangle and the smaller triangle

Area(larger triangle) = \(\frac{1}{2}*3*3 = \frac{9}{2}\)
Area(smaller triangle) = \(\frac{1}{2}*1*1 = \frac{1}{2}\)

Therefore, the area of the shaded region is \(\frac{9}{2} - \frac{1}{2} = \frac{8}{2} = 4\)(Option B)
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Re: In the coordinate plane above, if the equation of ℓ1 is y = x − 3 and [#permalink]

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New post 11 Dec 2017, 10:52
Is my approach right to get the value of Y?
as it is given that y = x − 3
and we know that Li intersect L2 at x = 2 then putting the value of x in above equation we can get the value of Y and that is -1...am i doing some mistake?
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Re: In the coordinate plane above, if the equation of ℓ1 is y = x − 3 and [#permalink]

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New post 30 Dec 2017, 22:56
sananoor wrote:
Is my approach right to get the value of Y?
as it is given that y = x − 3
and we know that Li intersect L2 at x = 2 then putting the value of x in above equation we can get the value of Y and that is -1...am i doing some mistake?

sananoor What your doing is perfectly correct. the co-ordinates in clockwise are (0,0) (2,0) (3,0) (2,-1) (0,-3)
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Re: In the coordinate plane above, if the equation of ℓ1 is y = x − 3 and   [#permalink] 30 Dec 2017, 22:56
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