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Question of the Week - 39 (In the x-y plane, PQ is a straight line)

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Question of the Week - 39 (In the x-y plane, PQ is a straight line)  [#permalink]

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New post 07 Mar 2019, 23:04
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e-GMAT Question of the Week #39

In the x-y plane, PQ is a straight line. The coordinate of point P is (a, b), where a > 0. Does point Q lie in the third quadrant?

    Statement 1: The origin bisects PQ in 1: 1 ratio.
    Statement 2: The slope m of the line PQ is non-negative.


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Re: Question of the Week - 39 (In the x-y plane, PQ is a straight line)  [#permalink]

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New post 07 Mar 2019, 23:18
IMO E

Both statements do not indicate anything. It could also be that the line can be just the X axis

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Re: Question of the Week - 39 (In the x-y plane, PQ is a straight line)  [#permalink]

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New post 08 Mar 2019, 03:46
IMO C

We know that P has a>0, so point P lie or in the first quadrant or in the fourth quadrant.

1) statment 1 not sufficient, because it tells us that the origin bisects the segment PQ in two equal segments. But Point Q can lie either in the second or in the third quadrant;

2) statment 2 not sufficent. We can only say that point Q cannot be in the second quadrant.

1+2) sufficient.

from 1: origin bisects PQ with ratio 1:1
from 2: the slope isn't negative, so Q can't be in the second quadrant.

IMO P lie in the first quadrant and Q in the third.

Hope my answer is correct.
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Re: Question of the Week - 39 (In the x-y plane, PQ is a straight line)  [#permalink]

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New post 10 Mar 2019, 07:11
EgmatQuantExpert wrote:
e-GMAT Question of the Week #39

In the x-y plane, PQ is a straight line. The coordinate of point P is (a, b), where a > 0. Does point Q lie in the third quadrant?

    Statement 1: The origin bisects PQ in 1: 1 ratio.
    Statement 2: The slope m of the line PQ is non-negative.


Image


1) Only tells us that x-axis will cut the line in 2, but still don't know anything about P and Q positions
2) Only tells us that the line is either horizontal or increasing

1)+2) still ambiguous about which one between P or Q is at the bottom or top of x-axis
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Re: Question of the Week - 39 (In the x-y plane, PQ is a straight line)  [#permalink]

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New post 13 Mar 2019, 01:19

Solution



Given:


    • PQ is a straight line.
    • Coordinate of point P is (a, b).
    • a > 0.

To find:

    • In which quadrant Q lies.

Let us assume the coordinate of point Q is (x, y)
To find the quadrant of point Q, we need to find the exact value of the coordinate of point Q.

Analysing statement 1:

The origin bisects PQ in 1: 1 ratio.
Hence, 0 = \(\frac{a + x}{2}\)and 0= \(\frac{b + y}{2}\)
• Therefore. x= -a and y= -b
• However, we still don’t know the exact value of x and y.

Hence, statement 1 is not sufficient to answer the question.

Analysing statement 2:

The slope m of the line PQ is non-negative.
Slope m of a line is \(\frac{y2 – y1}{x2-x1}\).
Hence, m= \(\frac{y-b}{x-a}\)
However, we still don’t have any information about the values of x and y.
Hence, statement 2 is not sufficient to answer the question.

Combining both the statements together:

From statement 1:
• x= -a and y= -b
From statement 2:
• m= \(\frac{y-b}{x-a}\)
We cannot find the value of x and y even after combining both the statements together.
Hence, option E is the correct answer.

Correct answer: option E.
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Re: Question of the Week - 39 (In the x-y plane, PQ is a straight line)  [#permalink]

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New post 15 Mar 2019, 04:37
EgmatQuantExpert Bunuel
EgmatQuantExpert wrote:

Solution



Given:


    • PQ is a straight line.
    • Coordinate of point P is (a, b).
    • a > 0.

To find:

    • In which quadrant Q lies.

Let us assume the coordinate of point Q is (x, y)
To find the quadrant of point Q, we need to find the exact value of the coordinate of point Q.

Analysing statement 1:

The origin bisects PQ in 1: 1 ratio.
Hence, 0 = \(\frac{a + x}{2}\)and 0= \(\frac{b + y}{2}\)
• Therefore. x= -a and y= -b
• However, we still don’t know the exact value of x and y.

Hence, statement 1 is not sufficient to answer the question.

Analysing statement 2:

The slope m of the line PQ is non-negative.
Slope m of a line is \(\frac{y2 – y1}{x2-x1}\).
Hence, m= \(\frac{y-b}{x-a}\)
However, we still don’t have any information about the values of x and y.
Hence, statement 2 is not sufficient to answer the question.

Combining both the statements together:

From statement 1:
• x= -a and y= -b
From statement 2:
• m= \(\frac{y-b}{x-a}\)
We cannot find the value of x and y even after combining both the statements together.
Hence, option E is the correct answer.


Correct answer: option E.




Can you let me know where I have gone wrong in my reasoning.

I thought C is the answer and my reasoning is given below.

Given a straight line PQ, with co-ordinates of P as (a,b) where a > 0

=> P is either on the 1st quadrant or on x axis or on the 4rth quadrant. Any pattern of line with positive, negative, infinite or zero slope can be drawn using points in these quadrants.

STATEMENT 1 : Origin bisects the line in the ratio 1 : 1

From the set of lines that could possibly be the answer we can eliminate all lines that doesnt pass through the origin.

As line passes through origin and a >0, line y axis and all lines with slope infinity (|| to y axis) can be eliminated.

If point P is in quadrant 1 it has to pass though quadrant 3 to satisfy this condition. This implies that the line will have a positive slope

If b is 0, then PQ is part of the x axis itself, and the point Q will be (-a,0) => Slope is 0 here

If the point P lies in 4rth quadrant, then Q must lie in the 2nd quadrant so that the origin bisects PQ. Here PQ will have negative slope.

HENCE STATEMENT 1 is not enough to answer the question

STATEMENT 2 : slope is non negative.

0 is neither positive nor negative. So I believe this statement essentially means slope is positive. We cannot narrow down the options here. The point Q could be in first quadrant itself

However, combining the 2 we can conclude that Q should pass through origin and should be in third quadrant. Hence both together can answer the question
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Re: Question of the Week - 39 (In the x-y plane, PQ is a straight line)   [#permalink] 15 Mar 2019, 04:37
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