In the xy-coordinate system, what is the slope of the line that goes through the origin and is equidistant from the two points P = (1, 11) and Q = (7, 7)?
2
2.25
2.50
2.75
3
2
2.25
2.50
2.75
3
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Source: Beat The GMAT — Problem Solving |
Thanks Sibbineni, you are absolutely right, i guess I misinterpreted the question.
I think the way you did is the right way.
Thanks.
I think the way you did is the right way.
Thanks.
Last edited by parallel_chase on Wed Aug 06, 2008 1:40 am, edited 1 time in total.
slope of the line that goes through the origin and is equidistant from the two points P = (1, 11) and Q = (7, 7)
The point equidistant from the points P and Q is mid point of P and Q
(x1+x2/2,y1+y2/2) is the midpoint formula---(1)
here x1=1 y1=11 and x2=7 and y2=7
substitute in (1) we have
(8/2,18/2)
(4,9)
so the slope between (0,0) and (4,9) and is
slope formula is (y2-y1)/(x2-x1)----(2)
here x1=0 y1=0 and x2=4 and y2=9
substitute in (2) we have
9/4===>2.25
IMO B
The point equidistant from the points P and Q is mid point of P and Q
(x1+x2/2,y1+y2/2) is the midpoint formula---(1)
here x1=1 y1=11 and x2=7 and y2=7
substitute in (1) we have
(8/2,18/2)
(4,9)
so the slope between (0,0) and (4,9) and is
slope formula is (y2-y1)/(x2-x1)----(2)
here x1=0 y1=0 and x2=4 and y2=9
substitute in (2) we have
9/4===>2.25
IMO B
The slope of the line always needs to be the same irrespective of what points on the line we take . So if P and Q are the points on a line , the slope should still be the same even without computing the mid-point
Am I missing something?
Am I missing something?
I have a question, if the line go through the origin and is parallel to the line PQ, isn't this line equidistant from the two points P and Q ?
Yes the slope of any two parallel lines will always be same.malolakrupa wrote:Even if we assume what pepeprepa says , the slope of 2 parallel lines should be the same right?
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It's either posted incorrectly or the question is just plain wrong - what's the source?malolakrupa wrote:Is the problem posted correctly?
The only way that a "line" can be equidistant from two points is if it's a perpendicular bisector of the line joining those two points. The slope of the line equidistant from (1,11) and (7,7) has the inverse negative slope of the line joining the two points. Since the line joining the two points has a slope of (11-7)/(1-7) = -(2/3), the slope of the line equidistant from the two points is 3/2. However, this line does NOT pass through the origin.
In fact, there's no line that goes through the origin and is equidistant from (1,11) and (7,7) - therefore there's no correct answer to this question (i.e. it's an impossible question).
For the given answer to be correct, the question should be:
"In the xy-coordinate system, what is the slope of the line that goes through the origin and bisects the line joining points P = (1, 11) and Q = (7, 7)?"
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This is exactly what i was thinking Ian.The ques seems wrong to me as well.Surprisingly It is Manhattan GMAT Test !
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Who is this Ian of whom you speak!ricky wrote:This is exactly what i was thinking Ian.The ques seems wrong to me as well.Surprisingly It is Manhattan GMAT Test !
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I'm guessing it's me.Stuart Kovinsky wrote:Who is this Ian of whom you speak!ricky wrote:This is exactly what i was thinking Ian.The ques seems wrong to me as well.Surprisingly It is Manhattan GMAT Test !
A line which is equidistant from two points does not need to be a perpendicular bisector. It only needs to be a bisector (edit- unless the line is parallel to the line joining the two points). Consider the following two points:Stuart Kovinsky wrote:
The only way that a "line" can be equidistant from two points is if it's a perpendicular bisector of the line joining those two points.
(-3, 1) and (3, -1)
The x-axis is clearly equidistant from both points (we measure distances at right angles, and the x-axis is one below the first point, and one above the second). But while the x-axis bisects the line joining (-3, 1) and (3, -1), it is not perpendicular to that line.
It's not difficult to see why we only need the midpoint in the given question, but it's more difficult to demonstrate without a diagram. Draw two points A and B in quadrant I of the coordinate plane. Connect A and B, and label the midpoint M. Draw the line containing the origin and M. To measure the distance from A to this line, we draw a right-angled triangle connecting A to the line, with hypotenuse AM. This triangle is exactly congruent to the right-angled triangle you get when you measure the distance from B to the line (the angles are equal, and AM=BM). We don't need a perpendicular bisector; we only need the midpoint.
As has been established above, the midpoint of (1, 11) and (7, 7) is (4, 9), and the slope of a line through both (0,0) and (4,9) is 9/4. There is still a problem with the question, however. A line which is parallel to the line connecting (1,11) and (7,7) will also be equidistant from both points. So there are two possible answers, 9/4 and -2/3, and the question is wrong to speak of the line which is equidistant from the points; it should ask for the slope of a line which is equidistant.
Last edited by Ian Stewart on Thu Aug 07, 2008 1:07 pm, edited 1 time in total.
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Yah, I know it's you, but I'm pretty sure that it was supposed to be me!Ian Stewart wrote:I'm guessing it's me.Stuart Kovinsky wrote:Who is this Ian of whom you speak!ricky wrote:This is exactly what i was thinking Ian.The ques seems wrong to me as well.Surprisingly It is Manhattan GMAT Test !
Stuart Kovinsky wrote:
The only way that a "line" can be equidistant from two points is if it's a perpendicular bisector of the line joining those two points.
I disagree. A point which is equidistant from two other points does not need to be a perpedicular bisector; however, a line can only be described as equidistant from two points if the entire line is equidistant, and that will only be the case if the line is a perpendicular bisector of the line joining the two points in question; otherwise, one point on the line may be equidistant but all of the other points on the line won't be.Ian Stewart wrote:A line which is equidistant from two points does not need to be a perpendicular bisector. It only needs to be a bisector.
(I didn't find anything online that provided a different definition of the concept.)
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