slope of the line

I found this maybe it will help:

https://mathforum.org/library/drmath/view/54426.html

I'm not too sure about my answer but I think that the answer is B.

The only way the line is going to be equidistant from Point P and Point Q is if it passes through the midpoint of those two points. The midpoint of points P and Q can be calculated by doing the following:

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

We already know that the line passes through the origin. So now we can calculate the slope since we have two points.

9-0/4-0=9/4 or 2.25

rahulg83 wrote: My doubt is if the line is equidistant from the mentioned points, does that means mid-point of the line segment joining the two points P and Q lies on the line passing through origin.

yes

rahulg83 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)?


OA later

Are you sure you typed question correctly, verbatim?
Because I think the meaning of the question changed...
If yes then, I agree with truplayer256

#your doubt regarding equidistant: Basically any point on the line, which is perpendicular to segment PQ is equidistant to P & Q. (that makes question ambiguous... )

Suppose the equation of line is ax+by+c=0,since line passes from origin, c=0 and the equation can be mx - y =0 or ax + by +c =0
where a = m,b=-1 and c=0, note here the slope is m
distance of a point(x1,y1) from a line ax+by+c=0 is
d = |(ax1+by1+c)|/sqrt(a^2+b^2) since the distance is equal from the points (1,11) and (7,7) the euqation to solve is
|m-11|=|7m+7|
which gives m = 2.25 hence answer is B.

aim-wsc wrote:

rahulg83 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)?


OA later

Are you sure you typed question correctly, verbatim?
Because I think the meaning of the question changed...
If yes then, I agree with truplayer256

#your doubt regarding equidistant: Basically any point on the line, which is perpendicular to segment PQ is equidistant to P & Q. (that makes question ambiguous... )

Actually what the question says is the line is equidistant from P and Q. That means the distance of line from P and that from Q is equal. This from nowhere means that mid-point of PQ lies on the given line. And the solution is given assuming this condition itself. Though deepak_free's solution is justified i feel...

P.S. Correct me please if i am wrong regarding the concept..
The answer indeed in 2.25..

what defines a line equidistant from 2 points?? Does that mean each and every point on the line is equidistant to P and Q. In that case the line has to be perpendicular to the line joining P and Q. The line thus formed will not go through the origin(slope = 2/3 one point is the mid point of P and Q (4,9)).

If the question says any one point on the line is equidistant to P and Q and that line passes through origin. We can have n number of lines satisfying that condition with different slopes. One possible line is the one that passes through the midpoint of P and Q.

This Question is confusing...

Thanks
raama

IMO

Distance from a point to a line is defined as the length of the segment that lies on another line which passes through that point and is perpendicular to the first line.

In this case, any line passing through the midpoint of PQ would be equidistant from P and Q and since the question asks for the one that passes through the origin, then we know its equation will be of the form:
y = ax and midpoint (4,9) belongs to the line

==> Slope a = y/x = 9/4 = 2.25


Cheers,
Sam

mygmat.2009 wrote:IMO

Distance from a point to a line is defined as the length of the segment that lies on another line which passes through that point and is perpendicular to the first line.

In this case, any line passing through the midpoint of PQ would be equidistant from P and Q and since the question asks for the one that passes through the origin, then we know its equation will be of the form:
y = ax and midpoint (4,9) belongs to the line

==> Slope a = y/x = 9/4 = 2.25


Cheers,
Sam

The line formed is not perpendicular to PQ if it passes thru origin and midpoint of P and Q.

Thanks
raama

Question is right as I have faced this Q in my test and got it right .

Here is the answer..

1. say line is y=mx+c, going through origin so (0,0) will satisfy the eq. so c=0 hence eq is y=mx.

Now a point equidistent from (1, 11) and (7, 7) will be ( (1+7)/2, (11+7)/2 )

so this point will satisfy y=mx
=> 9=4m => m= 2.25
Answer B.

krisraam, the line we are looking for is not perpendicular to PQ, I was referring to the line we draw to calculate the distance from a point to a line.

So, if point M is the mid-point of PQ, and O is the origin, then segment OM is not perpendicular to PQ, however, shortest distance from P to OM is the length of segment PP', where P' belongs to line OM and PP' perpendicular to OM. Same for Q, distance from Q to OM is QQ' , where Q' belongs to line OM and QQ' perpendicular to OM. Since line OM bisects PQ, then QQ' = PP'.

Do you agree?

Cheers,
Sam