Please help with locus problem

jimi

New member
Joined
Feb 1, 2005
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Before anyone complains about me posting this problem again I'd like to say that my earlier post was not answered fully.

Problem is as follows:


O-----------------------A
|
|
|
|
|
|
|
B|

'Draw accurately, on the diagram above, the locus of points which are the same distance from the line OA and the line OB'
(The lines OA and OB are the same length and they are perpendicular to each other).
 
Thanks for reply. However, I always understood the distance of a point from a line is the perpendicular distance so is there another line that satisfies the locus which is perpendicular to the bisector of angle O and goes through O

The problem is the ambiguity of the question.
Code:
    O....Y----Z
    .
    .
    .
    X
X is at (0,-4)
YZ goes from (4,0) to (8,0)
There is a distance from YZ (extended to O) to X but the distance to X from YZ (not extended) MIGHT be the distance from Y (the closest point on YZ) to X or
the distance to X from YZ MIGHT be the distance from the midpoint of YZ to X.
For your problem I think it is undefined unless you limit X to being on a perpendicular both lines somewhere in the square in the fourth quadrent, and the locus of eqidistant points is the line between (0,0) and (A,-B) which the first answer assumed. The bisector of <AOB in that quadrent.
With this definition there is only one line that satisfies the condition.
 
I agree. I also think the question is undefined. What exactly is meant by 'distance to the line'? I think the questioner means shortest distance to the line segment.
There's a similar question like this which is:

_________

'What is the locus of points 2 cm from the line above?'

The answer is given as:

_ _ _ _ _ _ _
_ _________ _
_ _ _ _ _ _ _

where the dashed lines represent the locus, i.e. 2 parallel lines 2 cm away and semi-circles at the ends.

The interpretation seems to be '2 cm away from the nearest point on the line segment' as the points on the semi-circles do not have a perpendicular distance of 2 cm from the line.
 
jimi said:
Before anyone complains about me posting this problem again I'd like to say that my earlier post was not answered fully.
You may not have gotten the answer you wanted, but that doesn't mean you didn't get an answer.

Eliz.
 
'You may not have gotten the answer you wanted, but that doesn't mean you didn't get an answer.

Eliz.'

Thank you for bringing that to my attention. As helpful as ever!
 
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