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In the coordinate plane above, if the equation of ℓ1 is y = x − 3 and
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03 Jan 2016, 12:28
Question Stats:
72% (01:59) correct 28% (02:13) wrong based on 138 sessions
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Re: In the coordinate plane above, if the equation of ℓ1 is y = x − 3 and
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04 Jan 2016, 04:37
Bunuel wrote: 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 Attachment: 20160104_0026.png I'll calculate this one using the area formula of the trapezoid. ℓ 2: x=2 means the hieght of the trapezoid is equal to 2 ℓ 1: y=x3, 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 ℓ 1&2 > y=23=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
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03 Jan 2016, 23:45
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
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07 Dec 2017, 11:11
When we reverse the diagram for simplicity, the triangles formed have the following coordinates Attachment:
Inverted diagram.png [ 2.67 KiB  Viewed 994 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
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11 Dec 2017, 09:52
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
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30 Dec 2017, 21:56
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 coordinates 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
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17 Feb 2019, 20:39
Bunuel wrote: 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 Attachment: 20160104_0026.png Identify all the boundary values From clockwise, x axis > 2,0 2,1 3,0 & 0,0 Put x = 0 in l1, to get y =3 Area of trapezium = 1/2 * 4 * 2 (sum of  sides * height * 1/2) = 4 B
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Re: In the coordinate plane above, if the equation of ℓ1 is y = x − 3 and
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