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3 Responses
A tidy way is to avoid slopes and use the line’s normal. Write the line as ax + by + c = 0. For a point (x0, y0), its reflection across that line is (x0 − 2a d, y0 − 2b d), where d = (a x0 + b y0 + c)/(a^2 + b^2). Here y = −x + 3 is x + y − 3 = 0, so a = 1, b = 1, c = −3. With (x0, y0) = (2, 5), we get d = (2 + 5 − 3)/(1^2 + 1^2) = 4/2 = 2, hence the reflection is (2 − 2·1·2, 5 − 2·1·2) = (−2, 1). As a check, the midpoint between (2, 5) and (−2, 1) is (0, 3), which lies on x + y = 3, and the connecting segment has slope 1, perpendicular to the line’s slope −1. Would you like a short derivation of the subtract-2d(a, b) rule, or want to try another point on the same line?
I used to turn these into a 20‑step mess too. I’d write out two equations, forget which slope is which, and somehow end up reflecting across the wrong line entirely… So here are two fast ways I keep on a sticky note when the line is y = −x + b.
Big picture:
– The line y = −x + 3 is the same as x + y = 3.
– A perpendicular to that has slope 1 (I sometimes accidentally keep −1… which would make it parallel, oops).
– Reflecting across x + y = constant is “swap and negate” after a tiny translation.
Method A (translate, flip, translate back)
This is the one my brain finds the least error-prone.
1) Rewrite the line as x + y = 3. Its “center point” (the point where it would meet the origin if we translated) is halfway along its intercepts: that’s (1.5, 1.5). I call this the hinge point.
2) Translate everything so that hinge point becomes the origin:
– New coordinates: X = x − 1.5, Y = y − 1.5.
– In these coordinates, the line becomes X + Y = 0, i.e. Y = −X.
3) Reflect across Y = −X. The reflection rule there is (X, Y) -> (−Y, −X).
– I used to think it was (Y, X) with a minus on just one of them, but that’s for a different mirror. Quick check: a point on Y = −X should stay put.
4) Translate back to original coordinates.
Do it for your point (2, 5):
– Translate: X = 2 − 1.5 = 0.5, Y = 5 − 1.5 = 3.5.
– Reflect: (X, Y) -> (−3.5, −0.5).
– Translate back: x = −3.5 + 1.5 = −2, y = −0.5 + 1.5 = 1.
So the reflection of (2, 5) across y = −x + 3 is (−2, 1).
Method B (a tiny vector formula)
There’s also a neat one‑line formula using a normal vector. For a line written as n · x = c, reflecting a point p gives
p’ = p − 2 (n · p − c) / ||n||² · n.
I always (always!) forget whether the normal should be (1, −1) here, but for x + y = 3, a correct normal is n = (1, 1), and c = 3. If I accidentally used (1, −1), I’d be reflecting in the wrong diagonal.
Plug in p = (2, 5):
– n · p = 1·2 + 1·5 = 7.
– n · p − c = 7 − 3 = 4.
– ||n||² = 1² + 1² = 2.
– Scale: 2 * 4 / 2 = 4.
– p’ = (2, 5) − 4(1, 1) = (−2, 1). Same answer.
Shortcut check (the “midpoint test”)
If Q is the foot of the perpendicular from your point to the line, then Q is the midpoint between the point and its reflection. That means:
– Slope of PQ should be 1 (perpendicular to slope −1). I once mistakenly wrote −1 here and got a parallel segment… which is a dead giveaway something’s off.
– Compute the perpendicular through (2, 5): y − 5 = 1(x − 2) → y = x + 3.
– Intersect with y = −x + 3: set x + 3 = −x + 3 → 2x = 0 → x = 0, y = 3. So Q = (0, 3).
– Reflect by doubling Q and subtracting P: P’ = 2Q − P = (0, 6) − (2, 5) = (−2, 1).
A tiny extra worked example
Reflect (0, 0) across y = −x + 3.
– Using Method A quickly: translate by (1.5, 1.5) → (−1.5, −1.5). Reflect across Y = −X: (−1.5, −1.5) -> (1.5, 1.5). Translate back: (3, 3).
– Quick midpoint check: the midpoint between (0, 0) and (3, 3) is (1.5, 1.5), which lies on x + y = 3. Looks good.
Bottom line
– Fast mental route: translate by (1.5, 1.5), swap-and-negate, translate back.
– Or use the dot‑product formula with n = (1, 1), c = 3.
Your answer: reflecting (2, 5) across y = −x + 3 gives (−2, 1).
I love this question! Reflecting across a tilted “mirror” feels like trying to high-five your own reflection on a slanted bathroom mirror-you want to go straight toward the glass (perpendicular), touch it, then go the same distance past it. The trick is to do that without getting tangled in slopes and midpoints.
Here are two fast ways-one is a super-quick hack for your exact line, and the other works for any slanted line.
1) Super-quick hack for lines of the form y = −x + k
If the mirror has slope −1 (like y = −x + k), there’s a lovely shortcut:
Reflect (x, y) across y = −x + k by sending it to (k − y, k − x).
Why this works: the line y = −x + k passes through the “center” (k/2, k/2). If you shift the world so that point is the origin, reflecting across y = −x just swaps and negates the coordinates, and shifting back condenses to (k − y, k − x). No 20-step algebra!
Your example
Reflect (2, 5) across y = −x + 3.
Use the hack with k = 3:
(2, 5) → (3 − 5, 3 − 2) = (−2, 1).
Quick sanity checks (optional but satisfying!):
– Midpoint of (2, 5) and (−2, 1) is (0, 3), which lies on x + y − 3 = 0 (i.e., y = −x + 3).
– The slope of the segment between the point and its reflection is 1, which is perpendicular to −1.
So the reflection of (2, 5) over y = −x + 3 is (−2, 1).
2) One-line formula that works for any line
For any line written as a x + b y + c = 0 and a point P = (x0, y0), the reflected point P′ is:
P′ = ( x0 − 2 a d, y0 − 2 b d ),
where d = (a x0 + b y0 + c)/(a^2 + b^2).
This is the “drop a perpendicular, then go the same distance past” in compact form: (a, b) is a normal vector to the line; d tells you how far you are from the line along that normal.
Applying it to your example (just to show it matches the hack)
Line y = −x + 3 is x + y − 3 = 0, so a = 1, b = 1, c = −3.
P = (2, 5). Compute d:
d = (1·2 + 1·5 − 3)/(1^2 + 1^2) = 4/2 = 2.
Then P′ = (2 − 2·1·2, 5 − 2·1·2) = (−2, 1). Same result!
Another simple worked example (non −1 slope)
Reflect P = (3, 4) across y = 2x + 1.
Write the line as 2x − y + 1 = 0, so a = 2, b = −1, c = 1.
d = (2·3 + (−1)·4 + 1)/(2^2 + (−1)^2) = (6 − 4 + 1)/5 = 3/5.
P′ = (3 − 2·2·(3/5), 4 − 2·(−1)·(3/5)) = (3 − 12/5, 4 + 6/5) = (3/5, 26/5).
Why I like these: in practice, I either use the tiny “swap-and-subtract” trick when the slope is −1, or I use the general formula-which is basically “dot with the normal, scale, subtract twice.” It’s fast and you never have to wrestle with perpendicular slopes or midpoints again.
Further reading (nice visuals and derivations):
– Brilliant: Reflecting a point in a line (derives the general formula and shows examples): https://brilliant.org/wiki/reflecting-a-point-in-a-line/
– Wikipedia: Reflection in a line in the plane (matrix view and formulas): https://en.wikipedia.org/wiki/Reflection_(mathematics)#In_a_line_in_the_plane
Final answer to your specific question:
The reflection of (2, 5) over y = −x + 3 is (−2, 1).