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# In the figure above,line segment OP has slope 1/2 and line segment PQ

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In the figure above,line segment OP has slope 1/2 and line segment PQ [#permalink]

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30 Jun 2016, 15:11
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OG Q 2017 New Question
[Reveal] Spoiler: OA

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Md. Abdur Rakib

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Sentence Correction-Collection of Ron Purewal's "elliptical construction/analogies" for SC Challenges

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Re: In the figure above,line segment OP has slope 1/2 and line segment PQ [#permalink]

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30 Jun 2016, 16:38
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AbdurRakib wrote:

OG Q 2017 New Question

Dear AbdurRakib
Another good question. I am happy to respond.

Statement #1: this gives us the length of OP, and we have the slope of PQ, but we have no idea how high Q goes. Because we don't know this, we can't compute the slope of OQ. This statement, alone and by itself, is insufficient.

Now, forget statement #1.

Statement #2: this gives us Q = (5,4). Well, we know O is at the origin, (0,0). We know this, because this is one thing we are allowed to assume from a coordinate geometry diagram: if a point looks like it is at the origin, and especially this point is given the name "O," then we can be sure that it is exactly at the origin. Thus, we know the coordinates of O & Q, so we can find the slope, m = 4/5. With this statement, we can directly compute an answer to the prompt question. This statement, alone and by itself, is sufficient.

First statement insufficient, second sufficient. Answer = (B)

Does all this make sense?
Mike
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Re: In the figure above,line segment OP has slope 1/2 and line segment PQ [#permalink]

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01 Jul 2016, 08:30
B

this is a good question.

1)insuff

2) Q co-ordinaes are given and O is origin.So slope can be found

B

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Re: In the figure above,line segment OP has slope 1/2 and line segment PQ [#permalink]

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05 Jul 2017, 06:49
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Re: In the figure above,line segment OP has slope 1/2 and line segment PQ [#permalink]

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05 Jul 2017, 07:32
AbdurRakib wrote:

OG Q 2017 New Question

Lets Assume P (a,b) and Q (x,y)
It is given that
Slope of OP
$$b/a=1/2$$
$$a=2b$$

Slope of OQ is $$y/x$$...Required..

Slope of PQ will be $$y-b/x-a = 2$$

y-b=2(x-2b)
y=2x-3b

(1)
OP = 2\sqrt{5}
OP = \sqrt{(a)^2 + (b^2)}
20 = a^2 + b^2
20 = 5b^2
b=+2 or -2
then a would be +4 or -4
If we use (4,2) to solve for (x,y) to reach to slope then
y-2/x-4 =2
y-2=2x-8
y=2x-6 We do not have another equation to solve for (x,y)

If we use (-4,-2) to solve for (x,y) to reach slope then
y+2/x+4 =2
y+2=2x+8
y=2x+6 We do not have another equation to solve for (x,y)

Hence Not Sufficient

(2) Q(5,4)--> x=5, y=4

y/x = 4/5

Bunuel Can u explain the approach, my wife is saying something is wrong
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Re: In the figure above,line segment OP has slope 1/2 and line segment PQ [#permalink]

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02 Sep 2017, 10:22
mikemcgarry why are we allowed to assume that O is at origin?

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Re: In the figure above,line segment OP has slope 1/2 and line segment PQ [#permalink]

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03 Sep 2017, 15:15
ambar09d wrote:
mikemcgarry why are we allowed to assume that O is at origin?

Dear ambar09d,

I'm happy to respond.

My friend, on GMAT PS geometry diagrams, there are lots of special cases that we can't assume---we can assume parallel or right angles or two equal sides simply because something looks that way. "Almost" doesn't count in geometry.

Nevertheless, there are a few basic assumptions we can make. If a line looks straight, we can assume that it is straight, that is has no hidden bend or curve to it. Also, if a point is portrayed as "on" a line, we can assume that it is in fact, "on" the line and not simply near the the line. In this diagram, point O is shown at the juncture of the x- and y-axes. We absolutely can assume that it is on both of those lines, and therefore, must be the intersection, the origin. Also, conventionally, the origin is often labeled O. For a problem to have a point labeled O near the origin but not exactly on it would a level of deviousness that we simply don't see on the GMAT.

Does all this make sense?
Mike
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Re: In the figure above,line segment OP has slope 1/2 and line segment PQ [#permalink]

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03 Sep 2017, 23:21
ambar09d wrote:
mikemcgarry why are we allowed to assume that O is at origin?

Well x-axis and y-axis only meet at 1 point that is Origin with co-ordinate (0,0).

By figure O is marked at intersection of 2 axis so its the origin.

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Re: In the figure above,line segment OP has slope 1/2 and line segment PQ [#permalink]

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03 Sep 2017, 23:27
Slope OP=1/2
Slope PQ=2
Find slope OQ
Assume p =(px,py)
Q= (qx,qy)
And as O is origin O= (0,0).

So slope OP py/px=1/2
slope PQ = (qy-py)/(qx-px) =2
Find slope qy/qx

1) it gives distance of OP =2root5
by distance formula px^2+py^2 =(2root5)^2
As py/px=1/2 => px=2py. Putting this value in above equation
=> 5py^2=20
=> py=2 => px=4

Putting these values in slope PQ => (qy-2)/(qx-4)=2
Not enough detail to find value of qx and qy or there ratios.

Not sufficient.

2) Direct co-ordinate of Q is given (5,4), so we cna find slope of line OQ.
Slope OQ = (5-0)/(4-0) = 5/4

Sufficient

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Re: In the figure above,line segment OP has slope 1/2 and line segment PQ   [#permalink] 03 Sep 2017, 23:27
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