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by aatech » Tue Jun 03, 2008 12:25 pm
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)?

a) 2
b) 2.25
c) 2.50
d) 2.75
e) 3
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by VP_RedSoxFan » Tue Jun 03, 2008 1:11 pm
We're trying to find information about the line coming out of the origin and going to the midpoint between P and Q.

In order to calculate the slope of a line is rise/run and since we're calculating the slope of a line coming out of the origin, the slope will simply be the y-coordinate/x-coordinate of the midpoint between P and Q.

Therefore, the challenging part of this question is finding the coordinates of the midpoint. Moving from P to Q goes down the x-axis 4 and over to the right 6. Therefore, the midpoint would move down from P by 2 and across 3 making the midpoint (4,9). (You can check this by moving again down 2 and across 3 to (7,7).)

From above, we know the slope is then 9/4, or 2.25.

Answer =

Hope this helps.
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by netigen » Tue Jun 03, 2008 1:12 pm
find the midpoint of P and Q = (4,9)

Two points on the line = (0,0) and (4,9)

slope = 9/4
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by aatech » Wed Jun 04, 2008 9:33 am
That's what I am not getting.. Why do we have to find the mid point of Points P and Q. The question is concerned about a point EQUIDISTANT from P and Q.. IMO, equidistant and midpoint are 2 different things. Although, considering it as midpoint is one possible solution...
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by netigen » Wed Jun 04, 2008 9:50 am
Isn't midpoint equidistant ?

If you know of a formula that can find a different equidistant point then you can use that too. All you have to do is to find one point that is equidistant.
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by AleksandrM » Wed Jun 04, 2008 9:53 am
Midpoint and equidistant IS the same thing. The middle point is a point that is midway between the two points P and Q. - 4/9 is also the the slope of a line perpendicular to the line discribed in the problem. This problem could have been made more difficult if the asked for the slope of a line perpendicular to the line described in the problem.
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by netigen » Wed Jun 04, 2008 10:30 am
AleksandrM wrote:Midpoint and equidistant IS the same thing. The middle point is a point that is midway between the two points P and Q. - 4/9 is also the the slope of a line perpendicular to the line discribed in the problem. This problem could have been made more difficult if the asked for the slope of a line perpendicular to the line described in the problem.
I beg to differ. Midpoint is just one of the possible equidistant points. All equidistant points will form a straight line in this case.
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by ksh » Wed Jun 04, 2008 11:57 pm
Just to add to netigen's words I would like to mention here that if a line passes through origin and the two equidistant points fall opposite to the line, the line passing trough PQ shall be normal to the line passing through origin.
Hence, m1*m2=-1

This is the condition of perpendicularity. If slope of PQ is known, slope of the line passing through origin can be calculated.
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Please clarify..

by bhatiagagan » Sat Jul 25, 2009 9:33 pm
Hi, i kinda stumbled upon this page while searching for this question.

I'd like to point out something here...
We all know that whenever we are calculating the distance of a point from a line, we take the perpendicular distance.
Now we also know, that there will be a line, parallel to the line throught these points, and passing through 0,0. this line will have a slope of -2/3.

What i dont get is, that if the line passing through the midpoint of these points is indeed the "line passing through 0,0 and equidistant from these points" shouldn't the slope of this line and the slope we found in the previous step follow -> m1m2= -1

I think the approach used is the wrong one, since a line passing through the midpoint of two given points and passing through 0,0 might not necessarily be perpendicular to the line through these points
Please Clarify
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by raghavsarathy » Sun Jul 26, 2009 3:51 am
We need to find a line which is equidistant to P and Q. This is only satisfied by a perpendicular bisector.

The midpoint of P and Q will definetly lie on this perpendicular bisector.

The midpoint of P and Q will (4,9)

This line also passes through the origin. We can obtain the slope since we have 2 points on the line. In this case .
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by singalong » Thu Jun 09, 2011 6:46 am
I am finding it difficult to understand this problem.Can someone gimme a graphical represntation?or paint me a rough picture?
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by singalong » Thu Jun 09, 2011 6:51 am
Is this it?
Attachments
midpoint slope.jpg
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by MBA.Aspirant » Thu Jun 09, 2011 7:19 pm
aatech wrote: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)?

a) 2
b) 2.25
c) 2.50
d) 2.75
e) 3
You got one point (0,0)

the other one is a mid-point so (1+7/2 , 7+11/2) (4,9)

slope = 9-0/4-0 = 9/4 = 3/2 = 2.5
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by peelamedu » Thu Jun 09, 2011 7:54 pm
Could someone explain the logic or formula to get the midpoint, given the x and y coordinates of two points.
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by MBA.Aspirant » Thu Jun 09, 2011 8:04 pm
Mid-point of (x,y) and (a,b) = (x+a/2 , y+b/2)


so in this example of P (1, 11) and Q (7, 7)

mid-point = (1+7/2, 11+7/2) = (4, 9)
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