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kiran120680
If P, Q(-3,4) and R(-4,-3) are 3 points in the coordinate plane, is QP>PR?

I. P is a point lying on the line y=x
II. Point P is in the first quadrant.



In no way Option A can be correct because
If P lies in 1st quadrant => QP<PR for all values of P
But
If P lies in 3rd quadrant => QP>PR for all values of P.

IMO, Option B is sufficient to say that QP>PR. Please correct the OA.
Thanks.
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kiran120680
If P, Q(-3,4) and R(-4,-3) are 3 points in the coordinate plane, is QP>PR?

I. P is a point lying on the line y=x
II. Point P is in the first quadrant.

I think what you intended to do here was to use y=x as the mirror but the values you provided for Q and R didn't work well.
Had the point been (-3, 4) and (4, -3), they would have been at the same distance from x=y and then if the point P was on x=y line, it would be at the same distance from both these points.
Maybe it's just a typo. It would have been a good question though!
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Can you check the OA? I don't believe this Statement (1) is sufficient.

If If P, Q(-3,4) and R(-4,-3) are 3 points in the coordinate plane, is QP>PR?

If P = (-3, -3), let's say, then PR = 1 and QP = 7, the answer is YES.
But if P = (4, 4), then PR =10.6 and QP = 7, then answer is NO.

Sorry for the typo error. Now corrected it.
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To do this question faster, instead of using distance formula to take out values of line segment QP and PR, better way would be to plot the points and randomly take based on statement 1 and statement 2 and
you'll see on graph that from statement 1 QP and PR are equal and hence we get a definite NO so statement 1 is SUFFICIENT.

BUT from statememt 2, when we plot points, it can be observed that there is no definite answer. There are two answers YES and NO. Therefore, NOT SUFFICIENT.

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Can someone please explain it in much detail? My coordinate geometry is not that great. I used a graph method for deciphering the question but somehow got confused midway

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kiran120680
If P, Q(-3,-4) and R(-4,-3) are 3 points in the coordinate plane, is QP>PR?

I. P is a point lying on the line y=x
II. Point P is in the first quadrant.



Please refer to the attach for quick approach on line y=x
Attachments

File comment: visual approach when we see y=x line
approach.jpeg
approach.jpeg [ 65.14 KiB | Viewed 5938 times ]

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