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A line segment is drawn in the xy-coordinate plane with endpoints P(3,

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A line segment is drawn in the xy-coordinate plane with endpoints P(3,  [#permalink]

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New post 29 Nov 2019, 03:26
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A line segment is drawn in the xy-coordinate plane with endpoints P(3,5) and Q(8,0). Line n is drawn so that it intersects PQ. What is the slope of line n?

(1) Point A is located on line n, and AP=AQ

(2) The x-intercept of line n is (3,0)


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A line segment is drawn in the xy-coordinate plane with endpoints P(3,  [#permalink]

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New post Updated on: 30 Nov 2019, 00:51
1
slope of given line is ; 3-8/5-0 ; -1
#1
Point A is located on line n, and AP=AQ
the slope of line will be 1 as it does intersect at 90*
sufficient
#2
The x-intercept of line n is (3,0)
again many possiblities of having lines which can intersect line PQ insufficient
insufficient
IMO A


A line segment is drawn in the xy-coordinate plane with endpoints P(3,5) and Q(8,0). Line n is drawn so that it intersects PQ. What is the slope of line n?

(1) Point A is located on line n, and AP=AQ

(2) The x-intercept of line n is (3,0)

Originally posted by Archit3110 on 29 Nov 2019, 03:59.
Last edited by Archit3110 on 30 Nov 2019, 00:51, edited 2 times in total.
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Re: A line segment is drawn in the xy-coordinate plane with endpoints P(3,  [#permalink]

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New post 29 Nov 2019, 04:56
I think explanation for statement- 2 needs review.
slope of n= 1 is conclusion of statement 1. slope(n) =1 is not mentioned in question stem.

For statement-1
A is on the perpendicular bisector of PQ.
Slope (PQ)= -1
then Slope (n)= 1
A is sufficient.

Statement-2
c= -3m
We don't know about c.
B is insufficient.

Archit3110 wrote:
slope of given line is ; 3-8/5-0 ; -1
#1
Point A is located on line n, and AP=AQ
the slope of line n must be 1 ; given point AP =AQ
so using distance formula we can determine point A ; sufficient
#2
we know the slope of line n = 1
y=mx+c
0=3+c
c=-3
eqn of line can be determined and slope m=1
sufficient
IMO D


A line segment is drawn in the xy-coordinate plane with endpoints P(3,5) and Q(8,0). Line n is drawn so that it intersects PQ. What is the slope of line n?

(1) Point A is located on line n, and AP=AQ

(2) The x-intercept of line n is (3,0)
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A line segment is drawn in the xy-coordinate plane with endpoints P(3,  [#permalink]

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New post Updated on: 29 Nov 2019, 21:06
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Given that a line is drawn from P(3,5) to Q(8,0). We know that line n intersects PQ, but exactly where line n intersects line PQ, we don't know. We are to determine the slope of line n.

Statement 1: Point A is located on line n, and AP=AQ
Statement 1 is insufficient. This is because from statement 1, line n can be a locus of points equidistant from points P and Q. In this case, line n is a perpendicular line to line PQ passing through the midpoint between the endpoints on line PQ, and the slope will be the negative reciprocal of the slope of line PQ.
But if by definition of point A on line n whereby AP=AQ implies that point A is the point of intersection between line n and line PQ, then we cannot conclusively determine a specific slope for line n. since n doesn't necessarily have to be perpendicular to line PQ for it to intersect line PQ at A such that AP=AQ.

Statement 2: The x-intercept of line n is (3,0).
Statement 2 is also insufficient. This is because we don't know the specific point where line n and line PQ intersect. So we cannot determine a specific slope for line n based on statement 2 alone.

1+2
Insufficient. Earlier on, I was a proponent for the view that both statements when taken together is sufficient. However, I had to change my mind on the sufficiency of combining 1 and 2. Why? If this question was well thought out as I believe it indeed was, statements 1 and 2 combined will still not be sufficient. We need to understand that the x-intercept of n (3,0) is also equidistant from P and Q.

To verify, let B denote the x-intercept of n. So B(3,0). Distance between P(3,5) and B(3,0)=√(0^2+5^2)=5.
Distance between B(3,0) and Q(8,0) = √((8-3)^2+0)=5.
What about if point A mentioned is actually the same point as the x-intercept? This is possible since the x-intercept of n is also equidistant from P and Q. Statement 1 is not definite on the specific location of point A on line n. So, A can actually be the x-intercept as given in statement 2. When this happens, then we cannot definitely say that n is perpendicular to PQ. As a result, statements 1 and 2 even when taken together are not sufficient.

The answer is, therefore E.

Originally posted by eakabuah on 29 Nov 2019, 05:08.
Last edited by eakabuah on 29 Nov 2019, 21:06, edited 1 time in total.
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A line segment is drawn in the xy-coordinate plane with endpoints P(3,  [#permalink]

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New post 29 Nov 2019, 05:09
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P could be on line AQ so slope zero or P could be on perpendicular bisector. Hence we can find it whatever it be.
Insuff

B

No direction no slope . Only one point cannot dictate the equation of line n
Insuff

C--- Eliminates the possibility of P on AQ. HENCE, C


Kudos if you liked my explanation.

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Re: A line segment is drawn in the xy-coordinate plane with endpoints P(3,  [#permalink]

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New post 29 Nov 2019, 09:03
(1) Point A is located on line n, and AP=AQ......... let (x,y) be the equidistant point from p,q...... Using distance formula for AP=AQ
(x-3)^2+(y-5)^2=(x-8)^2+y^2
x-y=3
The points which satisfies this equation....will be equidistant from p,q....this is the equation that intersects pq
Slope is 1
Sufficient

(2) The x-intercept of line n is (3,0)....... clearly insufficient....since with this we cannot find slope

OA:A

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A line segment is drawn in the xy-coordinate plane with endpoints P(3,  [#permalink]

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New post Updated on: 30 Nov 2019, 18:37
Slope of PQ = (0-5)/(8-3) = -5/5 = -1

(1) Point A is located on line n, and AP=AQ
Even though the point is equidistant from P and Q, it need not lie on the perpendicular bisector of PQ. So, slope is not fixed —>Insufficient


(2) The x-intercept of line n is (3,0)
We need to know at least 2 points on the line to find the slope —> Insufficient

Combining (1) & (2),
Note that (3,0) is equidistant from P and Q.
Also, point A is equidistant from P and Q.
So, the line passes through 2 points who are equidistant from end points of a line segment. Hence the line n is perpendicular bisector of PQ.

So, Slope of n*slope of PQ = -1
—> Slope of n*(-1) = -1
—> Slope of n = 1 —> Sufficient

IMO Option C

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Originally posted by Dillesh4096 on 29 Nov 2019, 18:40.
Last edited by Dillesh4096 on 30 Nov 2019, 18:37, edited 1 time in total.
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A line segment is drawn in the xy-coordinate plane with endpoints P(3,  [#permalink]

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New post 30 Nov 2019, 00:47
Bunuel wrote:

Competition Mode Question



A line segment is drawn in the xy-coordinate plane with endpoints P(3,5) and Q(8,0). Line n is drawn so that it intersects PQ. What is the slope of line n?

(1) Point A is located on line n, and AP=AQ

(2) The x-intercept of line n is (3,0)


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Easiest way to solve is to use graphical method(visualize).
Refer snapshot.
Attachment:
File comment: Linesegment PQ.JPG
Linesegment PQ.JPG
Linesegment PQ.JPG [ 93.57 KiB | Viewed 457 times ]

Statement 1: Red line is 'line n'. Dotted red lines are various cases possible in statement. Even though we have a number of cases we see that slope of line n remains same which is perpendicular to line segment PQ. And since we can calculate slope of PQ(-ve here), we can get slope of line n.
(\(Slope of PQ = \frac{rise}{run} = \frac{0-5}{8-3} = -1\); Grid height and width differ a little hence perpendicularity is affected, If you make it on grid-pad it would be clear)

SUFFICIENT.

Statement 2: Dotted Green lines indicate various possibilities of line n that have x-intercept of line n at (3,0).

INSUFFICIENT.

Answer A.
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Re: A line segment is drawn in the xy-coordinate plane with endpoints P(3,  [#permalink]

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New post 02 Dec 2019, 19:42
Bunuel
Please post the solution..
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Re: A line segment is drawn in the xy-coordinate plane with endpoints P(3,   [#permalink] 02 Dec 2019, 19:42
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