Find the equation of the line x - 2y + 1 = 0 when reflected about the line y = 2x - 3?

The line l': x - 2y + 1 = 0 has a slope k' = 1/2, the axis of symmetry s: y = 2x - 3 has a slope k = 2. Let l" (slope k " ) is symmetric image of l' with respect to s, then s is angle bisector of the angle between l' and l". Hence ∡(l', s) = ∡(s, l") in that order, because we deal with oriented angles. But

tan(∡(l', s)) = (k - k')/(1+ k'k) = tan(∡(s, l")) = (k " - k)/(1 + kk " )

The equation (2 - 1/2)/(1 + 2*(1/2)) = (k" - 2)/(1 + 2k ") yields k " = -5.5 and since l' and s have a common point (7/3, 5/3), we can write the equation of the required line:

l": y - 5/3 = k " (x - 7/3), finally l": 11x + 2y - 29 = 0.

But there is a shorter way without finding the common point: an angle bisector between 2 intersecting lines (any of 2 possible bisectors, which are perpendicular) is a locus of points, equidistant from them, so if

l' : a'x + b'y + c' = 0 and l" : a"x + b"y + c" = 0 are their equations, the equations of both bisectors are

| a'x + b'y + c' | / √(a' ² + b' ²) = | a"x + b"y + c" | / √(a"² + b"²)

/instead of absolute values you can take ± in either side/

You can use that if you know the equations of one of the lines and the bisector to find the equation of the other - convince yourself that simplifying the equation

-(x - 2y + 1) / √(1² + 2²) = (11x + 2y - 29)| / √(11² + 2²)

you get the given equation of s.