I’m struggling with reflections when the mirror line is slanted or doesn’t pass through the origin. I understand that a reflection should fix the mirror line and flip perpendicular distances, but I keep mixing up the perpendicular direction and the shift when the line isn’t at the origin.
For a concrete example: how do I reflect the point (2, 5) across the line y = x + 1? I’d also like to know how to reflect (4, 1) across the line y = -1/2 x + 2.
What is a reliable method here? I’m looking for a clear, step-by-step way (either a coordinate formula or a geometric construction) that works for any line y = mx + b. I also want a quick check I can use to verify that the reflected point is correct (e.g., something about perpendicularity and equal distances). I keep getting answers that don’t sit on the right perpendicular, so I think I’m missing a simple rule.
Please don’t derive anything heavy; I just want the method I should apply in these simple number cases and how to sanity-check the result.
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3 Responses
A clean, reliable recipe is this: first rewrite your mirror line as Ax + By + C = 0 (just move everything to one side). Then for a point P = (x0, y0), compute d = (A x0 + B y0 + C) / (A^2 + B^2). The reflected point is P’ = (x0 − 2 A d, y0 − 2 B d). Think of it like bouncing a ball off a wall: you go straight into the wall (perpendicularly), touch the wall, then head out the same distance on the other side.
Example 1: reflect (2, 5) across y = x + 1. That line is x − y + 1 = 0, so A = 1, B = −1, C = 1. d = (1·2 + (−1)·5 + 1)/(1^2 + (−1)^2) = (2 − 5 + 1)/2 = −1. Then P’ = (2 − 2·1·(−1), 5 − 2·(−1)·(−1)) = (4, 3). Quick checks: the midpoint of (2,5) and (4,3) is (3,4), which lies on y = x + 1; and the segment from (2,5) to (4,3) has slope −1, which is perpendicular to the mirror’s slope 1.
Example 2: reflect (4, 1) across y = −(1/2)x + 2. Write it as −x − 2y + 4 = 0 (any nonzero multiple is fine), so A = −1, B = −2, C = 4. d = (−1·4 + (−2)·1 + 4)/(1 + 4) = (−4 − 2 + 4)/5 = −2/5. Then P’ = (4 − 2·(−1)·(−2/5), 1 − 2·(−2)·(−2/5)) = (3.2, −0.6). Checks: midpoint is (3.6, 0.2), which satisfies y = −(1/2)·3.6 + 2 = 0.2; and the segment’s slope is 2, the negative reciprocal of −1/2.
Here’s the quick, clean rule I use when the mirror is y = m x + c (I always forget signs unless I do it this way): for a point P = (x0, y0), compute d = (m x0 − y0 + c)/(m^2 + 1). Then the reflection P′ is (x0 − 2 m d, y0 + 2 d). That’s it. It works because d is the signed perpendicular distance scaled by m^2+1, so you’re sliding exactly twice along the perpendicular. Sanity-checks: (i) the midpoint of P and P′ should lie on the line, and (ii) the segment PP′ should be perpendicular to the line (slope product −1). Now your examples: across y = x + 1 (so m = 1, c = 1) for P = (2, 5), d = (1·2 − 5 + 1)/2 = −1, so P′ = (2 − 2·1·(−1), 5 + 2·(−1)) = (4, 3); midpoint (3, 4) lies on y = x + 1 and the slope from (2,5) to (4,3) is −1, perpendicular to slope 1. Across y = −(1/2)x + 2 (m = −1/2, c = 2) for P = (4, 1), d = ((−1/2)·4 − 1 + 2)/(1/4 + 1) = −4/5, so P′ = (4 − 2·(−1/2)·(−4/5), 1 + 2·(−4/5)) = (16/5, −3/5); midpoint (18/5, 1/5) lies on the line and the slope 2 is perpendicular to −1/2.
My go-to “clean” method is: write the line in standard form A x + B y + C = 0, then reflect P = (x, y) by P’ = (x, y) − 2(Ax + By + C)/(A^2 + B^2) · (A, B). It’s like dropping a perpendicular “shadow” from your point onto the mirror, then walking the same distance through the mirror to the other side. For y = m x + b, just take A = m, B = −1, C = b. Example 1: reflect (2, 5) across y = x + 1, so A = 1, B = −1, C = 1. Compute s = Ax + By + C = 2 − 5 + 1 = −2 and A^2 + B^2 = 2, so P’ = (2, 5) − 2(−2/2)(1, −1) = (4, 3). Quick check: the segment from (2, 5) to (4, 3) has direction (2, −2), perpendicular to the line (slope −1 vs slope 1), and the midpoint (3, 4) lies on y = x + 1. Example 2: reflect (4, 1) across y = −(1/2)x + 2, so A = −1/2, B = −1, C = 2 (or scale to A = −1, B = −2, C = 4). Then s = −4 − 2 + 4 = −2 and A^2 + B^2 = 5, so P’ = (4, 1) − 2(−2/5)(−1, −2) = (16/5, −3/5). Checks: the displacement is (−4/5, −8/5) with slope 2, perpendicular to the line’s slope −1/2, and the midpoint (18/5, 1/5) sits on the line. If both perpendicularity and “midpoint on the mirror” pass, your reflection is golden.