Symmetric of a point relative to a line

AI Thread Summary
To find the symmetrical of a point relative to an arbitrary line, the process involves determining the perpendicular line from the point to the given line and calculating the distance from the point to the line. The distance can be computed using the formula d = |Ax + By + C| / (A^2 + B^2)^(1/2). While this method is effective, it can feel cumbersome when applied multiple times in a problem. Some participants express a desire for a quicker method to achieve the same result. Ultimately, using the distance formula once can yield the shortest distance necessary for finding the symmetrical point.
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Homework Statement


What is the easiest way of finding the symmetrical of a point relative to an arbitrary line?
(I was asked on an exam to find the symmetrical of a point relative to the line y = x, but that's rather trivial - just switch the coordinates. How can I do it for any arbitrary line ax + by = c?)

Homework Equations





The Attempt at a Solution


I found a way, but it's rather messy. Given the line s and the point P, find the line t that is perpendicular to s and passes through P. Calculate the distance d between P and s. Find the point on the line t that is also at a distance d from s.

This sounds awfully complicated and messy for me. Is there a quicker way?
 
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That's the way I'd do it. I don't know of any other.
 
For a line of the form Ax+By+C = 0, distance d from a point p is:

d = |Ax+By+C|/(A2+B2)1/2

[|...| is simply the absolute value meant to keep d>/=0]
 
Hitchens said:
For a line of the form Ax+By+C = 0, distance d from a point p is:

d = |Ax+By+C|/(A2+B2)1/2

[|...| is simply the absolute value meant to keep d>/=0]

This is precisely the formula I would use in the description I have given. But it's rather
cumbersome to use it twice in a single problem. Is there a way around?
 
Acut said:
This is precisely the formula I would use in the description I have given. But it's rather
cumbersome to use it twice in a single problem. Is there a way around?

Why would you use it twice? Using it once will provide you with the shortest distance from the line to the point.
 
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