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Line equidistant from two points

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by madhukumar_v » Wed Aug 04, 2010 7:08 pm
Line: y=mx+c
Going through origin, means c=0; so y=mx
now this line, y=mx is equidistant from two points. The distance between a line, ax+by+c and point (x1,y1) is given by

|a(x1)+b(y1)+c|/ sqrt(a^2+b^2)

so,
distance between y=mx and (1,11) = |m-11|/sqrt(m^2+1) ---(1)
distance between y=mx and (7,7) = |7m-7|/sqrt(m^2+1) --- (2)

since (1) = (2)
so, m-11= 7m-7
hence, 6m=-4; m= -2/3;
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by sheelanadh » Thu Aug 05, 2010 3:47 am
madhukumar_v wrote:Line: y=mx+c
Going through origin, means c=0; so y=mx
now this line, y=mx is equidistant from two points. The distance between a line, ax+by+c and point (x1,y1) is given by

|a(x1)+b(y1)+c|/ sqrt(a^2+b^2)

so,
distance between y=mx and (1,11) = |m-11|/sqrt(m^2+1) ---(1)
distance between y=mx and (7,7) = |7m-7|/sqrt(m^2+1) --- (2)

since (1) = (2)
so, m-11= 7m-7
hence, 6m=-4; m= -2/3;
Why not m-11= -(7m-7)?

Am i Missing something?
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by viju9162 » Thu Aug 05, 2010 7:44 am
IMO [spoiler]9/4[/spoiler]

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"Native of" is used for a individual while "Native to" is used for a large group
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by kvcpk » Thu Aug 05, 2010 9:40 am
gmatrant 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)
The formula that gives the distance between a point (m, n) and a line ax + by + c = 0 is:
d = |am+bn+c|/sqrt(a^2+b^2)
Line passing thru origin will have eqtn mx-y=0 and point is (1,11)

Hence d = |m-11|/sqrt(m^2+1)
Similarly, distance from (7,7) is
d= |7m-7|/sqrt(m^2+1)

Now both are equal.
Hence
|m-11|/sqrt(m^2+1) = |7m-7|/sqrt(m^2+1)

Now three cases arise
m>=11, 1<=m<11,m<1

When m>=11
(m-11)/sqrt(m^2+1) = (7m-7)/sqrt(m^2+1)
m-11 = 7m-7 -> m=-2/3 .. This is not greater than 11.. Hence Not Possible

When 1<=m<11,
(11-m)/sqrt(m^2+1) = (7m-7)/sqrt(m^2+1)
11-m = 7m-7
m=9/4

When m<1,
11-m = 7-7m
m=-2/3

I believe both values are possible. Whichever value is in the options can be chosen.

Hope this helps!!
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by commonsense27 » Thu Aug 05, 2010 3:08 pm
You guys are not solving it right.

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)

Midpoint (M) between P = (1,11) and Q = (7,7) will be (1+7)/2 and (11+7)/2 =(4,9)

so the slope of the line passing through origin and (4,9) = (9-0)/(4-0) = 9/4.
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by nox104 » Thu Aug 05, 2010 5:39 pm
Agree with commonsense27.

Slope of a line through 2 points = m = (y2 - y1) / (x2 - x1)
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by gmatrant » Fri Aug 06, 2010 5:01 am
yes Ans is 9/4
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by jmb » Wed Feb 06, 2013 5:07 pm
Seems to me that previous answers have missed the point (sic) - either the original question is malformed, or "none" is a permitted answer:

For a line to be equidistant to two points, every point along that line needs to be equidistant from them. Fundamental Pythagoras (1^2+11^2 > 7^2+7^2) shows you immediately that the two points given are not equidistant from the origin - therefore there is no line which both goes through the origin and is equidistant from the two points.

(4,9) is indeed the closest equidistant point, but it doesn't follow that this is the slope of the line. The slope of the line between the given points is -4/6, so the slope of the equidistant (perpendicular) line is 6/4; the equation for the equidistant line is y=1.5x+3. This crosses the y axis at (0,3) which you can readily calculate is also equidistant from the two points (and is plainly not the origin).
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by Tommy Wallach » Wed Feb 06, 2013 8:40 pm
Hey Guys,

Jmb is absolutely right here. A line can't be equidistant from two points. It isn't actually possible.

Some people are solving this question: "What is the slope of the line that goes through the origin and the midpoint of line segment PQ, where P = (1,11) and Q = (7,7). But that is not the question that's been asked. Furthermore, even a slight change of the question doesn't help: "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). This would have an infinitude of answers, as there are infinite points that would be equidistant from both P and Q.

And of course, if the question means to say that (1,11) and (7,7) are equidistant from the origin, that isn't possible either.

-t
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by Navinag » Thu May 30, 2013 11:40 am
Tommy Wallach wrote:Hey Guys,

Jmb is absolutely right here. A line can't be equidistant from two points. It isn't actually possible.

Some people are solving this question: "What is the slope of the line that goes through the origin and the midpoint of line segment PQ, where P = (1,11) and Q = (7,7). But that is not the question that's been asked. Furthermore, even a slight change of the question doesn't help: "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). This would have an infinitude of answers, as there are infinite points that would be equidistant from both P and Q.

And of course, if the question means to say that (1,11) and (7,7) are equidistant from the origin, that isn't possible either.

-t

Well.. the way I understand is, whenever the question asks about the 'distance between a line and a point', it refers to the shortest distance, which will be the length of the line drawn from the point to the line forming a right angle.

If that's how everybody else understood, then they started it right.


@commonsense27 for his method of calculation: The question doesn't say if both the points are on same side of the line or opposite sides. Even if the points are on opposite sides, by definition of 'distance from the line', your solution of finding the midpoint will not hold true.
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