✔ 最佳答案
In two dimensions:
If the line of reflection is ax+by=c
and the line to be reflected is dx+ey=f,
then one way to write the reflected line is
.. (-a^2 d + b^2 d - 2 a b e)x + (-2 a b d + a^2 e - b^2 e)y = -2 a c d + a^2 f - b (2 c e - b f)
or
.. (d(b^2-a^2) -2abe) x + (e(a^2-b^2) -2abd) y = f(a^2+b^2) -2c(ad+be)
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Example:
We want to reflect the line x+2y=4 about the line 5x+4y=3. The result is 89x + 22y = -86. The source link shows plots of these lines.
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The method used here is to have point (p, q) satisfy dp + eq = f and the midpoint of (p, q) and (r, s) fall on the line of reflection. Then we also have bp - aq = br - as, so that the segment between (p, q) and (r, s) is perpendicular to the line of reflection. Solving these equations for (p, q), we used those values in the equation of the original line (to be reflected) to find the relationship between r and s. The algebra is tedious, but not difficult. In the process, we ended up with a^2+b^2 in the denominator a number of places, so we multiplied by that to have "integer" coefficients.