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On the xy-coordinate plane, a quadrilateral is bounded by

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On the xy-coordinate plane, a quadrilateral is bounded by [#permalink] New post 17 May 2011, 03:07
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On the xy-coordinate plane, a quadrilateral is bounded by the x-axis, y-axis, the line y = 6, and the line y = ax + 8. What is the area of this region?

(1) The point (1,6) is on the boundary of the quadrilateral.

(2) The point (6,5) is on the boundary of the quadrilateral.

Attachment:
screen_shot_2011_03_11_at_3.28.58_pm.png
screen_shot_2011_03_11_at_3.28.58_pm.png [ 18.66 KiB | Viewed 3584 times ]


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[Reveal] Spoiler: OE
There are a few inferences we can start with before we turn to the statements. Since the lines y = 6 and the x-axis are parallel to one another (and the y-axis and the line y = ax + 8 are not parallel), this area is a trapezoid. Furthermore, since the x-axis and y-axis are parallel, this area is a right trapezoid (a trapezoid with two right angles). It is relatively easy to find the area of a right trapezoid; they can be divided into a rectangle and a triangle. All we need is the value of a. Then we will have the equations of all of the lines that form this trapezoid. It is neither advisable nor necessary to actually calculate the area of this trapezoid. It is better just to determine what information we will need to do so. In this case, we just need to find the equation of that fourth line. Now, turning to the statements:

(1) Remember that the line y = 6 is one of the borders of the trapezoid. This line is the horizontal line that passes through (1,6). Therefore, we don't know if the point given is on y = 6 or the vertex intersection with the line y = ax + 8. INSUFFICIENT

(2) The point (6,5) does not lie on the x-axis, the y-axis, or the line y = 6. Thus, it must lie on the line y = ax + 8. Substituting x and y into the equation we have:

5 = a(6) + 8
-3 = 6a
-0.5 = a

The equation of the line is y = -0.5x + 8

Now that we have this equation, we can solve the questions. SUFFICIENT.

Although we don't want to waste time finding the area, here's how we would do it for review purposes: the line we found crosses the x-axis and the line y = -0.5x + 8. Let's first find the intersection points.

6 = -0.5x + 8
-2 = -0.5x
x = 4

This is where the line we found meets the line y = 6. Draw an imaginary line from this point to the x-axis. this will split the trapezoid into a rectangle and a triangle. The area of the rectangle is 24 (4 wide x 6 long). Now let's find the other intersection point, the x-intercept of y = -0.5x + 8:

0 = -0.5x + 8
-8 = -0.5x
16 = x

So we have a right triangle with a base of 12 {from our imaginary line at (4,0) to the x-intercept at (16,0)} and a height of 6. Plugging these values into the formula for the area of a triangle, we get:

0.5(12 x 6) = 0.5(72) = 36

Adding up the two areas gives us a final value of 60.

The credited response is (B): Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
[Reveal] Spoiler: OA
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Re: Area of trapezium using coordinates [#permalink] New post 17 May 2011, 04:09
See the attached images. All co-ordinates mentioned can be found out by substituting correct co-ordinates.

Stmt1: The point (1,6) is on the boundary of the quadrilateral.
That means point is on line y=6. But we don't know the value of a. Insufficient.

Stmt2:The point (6,5) is on the boundary of the quadrilateral.
That means point is on y=ax+8. Substituting (6,5)
5=6a+8
a=-1/2

We know the value of a. We can find all the co-ordinates with a in attached image.
(-8/a,0)=> 16,0
(-2/a,6)=> 4,0

Area of shaded region will be bigger triangle minus smaller triangle.

Area of bigger triangle= 1/2 * 16 * 8 =64
Area of smaller triangle=1/2 * 4 * 2 =4

Area of shaded region= 64-4=60.

OA B.
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soln5.JPG
soln5.JPG [ 10.3 KiB | Viewed 3552 times ]


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Re: Area of trapezium using coordinates [#permalink] New post 17 May 2011, 08:00
a (1,6) lies on y= ax + 8 or y=6 as 6= ax + 8 means ax = -2.

Hence not sufficient.

b (6,5) on y = ax+8 gives 5 = 6a + 8 giving a = -1/2

also ax= -2 means x = 4

means y= ax + 8 when y = 0 = -1/2 x + 8 thus x = 16

thus area of trapezium = 1/2 * altitude * (sum of parallel sides )
= 1/2 * 6 * (16 + 4) = 60 Sufficient.

Hence B.
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Re: Area of trapezium using coordinates [#permalink] New post 17 May 2011, 09:51
3 sides are fixed with 3 lines
1. X axis
2. Y axis
3. y=6

Forth side need to be fixed. We have a equation of 4th line with one variable.
Anything that giving 4th equation will be sufficent to ans the q.

statement 2 give a point on that line and we can find the 4th line equation as well.
Hence B
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Re: Area of trapezium using coordinates [#permalink] New post 17 May 2011, 20:15
We need (Parallel Side 1 + Parallel Side2)/2 * Distance

Distance = 6

y = ax + 8 intersects x-axis at x = -8/a ( y = 0) and the line y = 6 at (-2/a,6)

So area = -10/a * 1/2 * 6




(1)

(1,6) can lies on y = 6, but it may lie on y = ax + 8 in a nonverlapping manner,i.e, it may be common point to both or may not be so.

InSufficient

(2)

Point (6,5) can lie only on y = ax + 8

=> 5 = 6x + 8

=> a = -3/6 = -1/2

Sufficient

Answer - B
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Re: Area of trapezium using coordinates [#permalink] New post 09 Jun 2011, 06:31
for 1)
point (1,6) lies on y= ax + 8 or 6= a.1 + 8 means a = -2.

Hence sufficient.

2) is already sufficient as proved earlier.

So my ans is D.
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Re: Area of trapezium using coordinates [#permalink] New post 09 Jun 2011, 07:30
I also agree with toughmat.
At Coordinate (1,6)=> x=1 and y=6
Now we also know that y=mx+c
According to question y=ax+8, by putting x=1 and y= 6 wecan get the value of a = -2, thus slope is -2.
Now we can calculate area of quadrilateral which comes to be 18.
Same in the case for second option. so, my answer is D.
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Re: Area of trapezium using coordinates [#permalink] New post 09 Jun 2011, 07:33
Hi Fluke/Subhash,
Can you explain me statement 1 in a better way .
if points (1,6) boundaries.. then why can't it lie on y =ax+8 ; 6 =a*1+8 a=-2?
or is it that we are given another y=6 ( which we know has to be a horizontal line parallel to x axis) and hence (1,6) can't lie on y = ax+8
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Re: Area of trapezium using coordinates [#permalink] New post 09 Jun 2011, 07:50
sudhir18n wrote:
Hi Fluke/Subhash,
Can you explain me statement 1 in a better way .
if points (1,6) boundaries.. then why can't it lie on y =ax+8 ; 6 =a*1+8 a=-2?
or is it that we are given another y=6 ( which we know has to be a horizontal line parallel to x axis) and hence (1,6) can't lie on y = ax+8


Please check the OE and see whether it helps.
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Re: Area of trapezium using coordinates [#permalink] New post 09 Jun 2011, 20:45
I don't understand why the OA is B

as in a trapezium formed by 4 lines so each line has to intersect at 2 points

we know that (0,0) and (6,0) are intersected by the x-axis, y-axis and the line y = 6 so the line y = ax + 8, y=6 and x-axis intersect at 2 other points.

so if the 3rd vertex is given as (1,6) then the line has to definitely pass through the intersection of y=6 and y = ax+8 how can we assume that the line intersects y-axis in a non contributing way? Can someone explain that with the help of a diagram if thats the case?

My reasoning is that each pairs of lines have unique intersection points and each point represents a vertex of the quadrilateral. this can be confirmed by the diagrams given below.

Sorry in the diagram below the slopes have to be -ve and +ve respectively.
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Re: Area of trapezium using coordinates [#permalink] New post 09 Jun 2011, 20:54
Are you by any chance talking about the trapezium in which the 4 points lie within the bounded region??

and if that were the case the answer should be E because we can only find the bounds of the bounded region and not the specific points in that region.
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Re: Area of trapezium using coordinates [#permalink] New post 10 Jun 2011, 10:55
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phanideepak wrote:
Are you by any chance talking about the trapezium in which the 4 points lie within the bounded region??

and if that were the case the answer should be E because we can only find the bounds of the bounded region and not the specific points in that region.


The question and solution are fine.

You have four lines: x axis, y axis, y = 6 and y = ax + 8
You need the value of a to get the area bounded by these lines.

Stmnt1: This tells you that (1, 6) lies on the boundary (one of the lines) of the quadrilateral (not that it is a vertex necessarily). We know that y = 6 passes through this point anyway. All this says is that y = 6 intersects with y = ax + 8 at a point (m, 6) such that m is greater than 1. We still do not know the value of a and hence this is not sufficient.

Stmnt2: None of x axis, y axis and y = 6 can pass through (6, 5). If (6, 5) lies on the boundary of the quadrilateral, y = ax + 8 has to pass through it. So we know a point on y = ax + 8 which will give us the value of 'a' and hence the equation of the fourth line. This will give us the area of the quadrilateral. Sufficient

Also, in your figure, the second diagram is not possible because that line cannot pass through (6, 5). Its y co-ordinate will be greater than 8 in the first quadrant.
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Re: Area of trapezium using coordinates [#permalink] New post 10 Jun 2011, 20:16
VeritasPrepKarishma wrote:
phanideepak wrote:
Are you by any chance talking about the trapezium in which the 4 points lie within the bounded region??

and if that were the case the answer should be E because we can only find the bounds of the bounded region and not the specific points in that region.


The question and solution are fine.

You have four lines: x axis, y axis, y = 6 and y = ax + 8
You need the value of a to get the area bounded by these lines.

Stmnt1: This tells you that (1, 6) lies on the boundary (one of the lines) of the quadrilateral (not that it is a vertex necessarily). We know that y = 6 passes through this point anyway. All this says is that y = 6 intersects with y = ax + 8 at a point (m, 6) such that m is greater than 1. We still do not know the value of a and hence this is not sufficient.

Stmnt2: None of x axis, y axis and y = 6 can pass through (6, 5). If (6, 5) lies on the boundary of the quadrilateral, y = ax + 8 has to pass through it. So we know a point on y = ax + 8 which will give us the value of 'a' and hence the equation of the fourth line. This will give us the area of the quadrilateral. Sufficient

Also, in your figure, the second diagram is not possible because that line cannot pass through (6, 5). Its y co-ordinate will be greater than 8 in the first quadrant.


Ah! Now i understand... I was assuming it be the vertex. Thanks a lot for your solution :) I need to improve my reading skills :(
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Re: On the xy-coordinate plane, a quadrilateral is bounded by [#permalink] New post 10 May 2013, 10:02
The mistake I did was substitute (1,6) in the equation y=ax+b. I got Ans: D. But each statement gives different slopes and intercepts, so I am hesitant to keep answer D. I think in GMAT if the option is D, the answer will be same.

The point (1,6) may be on line y=6.

I think this is the trap. Otherwise this question is easy.
Re: On the xy-coordinate plane, a quadrilateral is bounded by   [#permalink] 10 May 2013, 10:02
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