I’m trying to reflect the point (3, -1) across the line y = 2x + 1, and my brain keeps short-circuiting. With the x-axis or the line y = x I know the little tricks, but as soon as the “mirror” is tilted and doesn’t go through the origin, I’m lost. Do I have to draw a perpendicular to the line and find the midpoint, or is there a standard coordinate method that gets me the reflected point directly? I feel silly because I’m mixing up perpendicular slopes and the intercept and second-guessing every step.
Is there a straightforward way to get the coordinates without graph paper-like a formula or a reliable procedure I can do on a test? Do I need to first shift/rotate the line to make it easier and then undo that, or is there a direct approach?
Quick follow-up: if I’m reflecting a whole triangle over y = 2x + 1, is it always safe to reflect each vertex and then connect them, or is there something about orientation I should watch out for?
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3 Responses
There’s a one-shot formula-no graph paper, no rotating the whole world. Put the line in ax + by + c = 0 form. Then for a point (x0, y0), the reflection across that line is
(x’, y’) = (x0 – 2a(ax0 + by0 + c)/(a^2 + b^2), y0 – 2b(ax0 + by0 + c)/(a^2 + b^2)).
It’s just “subtract twice the signed distance in the normal direction,” so you never have to mess with perpendicular slopes.
Your line y = 2x + 1 is 2x – y + 1 = 0, so a = 2, b = -1, c = 1. For (3, -1): ax0 + by0 + c = 2*3 + (-1)*(-1) + 1 = 8, and a^2 + b^2 = 5. Plug in:
x’ = 3 – 2*2*(8/5) = 3 – 32/5 = -17/5,
y’ = -1 – 2*(-1)*(8/5) = -1 + 16/5 = 11/5.
So the reflected point is (-17/5, 11/5). Quick check: the midpoint lands on y = 2x + 1, so it’s right. For a whole triangle, yes-reflect each vertex and connect them in the same order. The shape stays congruent, but orientation flips (clockwise becomes counterclockwise). That’s normal for reflections.
Want a 30-second practice: pick another point, say (0, 0), and try reflecting it across the same line-what do you get?
Yes-there’s a direct, reliable formula. Write the mirror line in the form ax + by + c = 0. For y = 2x + 1 this is 2x − y + 1 = 0, so a = 2, b = −1, c = 1. If P = (x0, y0), then its reflection across that line is
x’ = x0 − 2a(ax0 + by0 + c)/(a^2 + b^2), y’ = y0 − 2b(ax0 + by0 + c)/(a^2 + b^2).
Plugging P = (3, −1): s = a x0 + b y0 + c = 2·3 + (−1)(−1) + 1 = 8 and a^2 + b^2 = 5, so
x’ = 3 − 2·2·8/5 = −17/5 and y’ = −1 − 2·(−1)·8/5 = 11/5. That’s the reflected point: (−17/5, 11/5). No rotation or shifting is needed; this is just the algebraic version of “drop a perpendicular to the line and go the same distance past it.” Like walking straight to a tilted mirror, tapping it, and taking the same number of steps further in the same straight line. A derivation of the formula is here: https://www.geeksforgeeks.org/reflection-of-a-point-about-a-line/
For a whole triangle, reflect each vertex with the same formula and then connect them. The reflected triangle is congruent to the original and lies where you expect. The only thing to note is orientation: reflections reverse it (clockwise becomes counterclockwise), which is normal and not a problem unless a question is tracking orientation explicitly.
Use the direct formula for reflecting across ax+by+c=0: (x’, y’) = (x, y) − 2(ax+by+c)/(a^2+b^2) · (a, b); for y=2x+1 (i.e., 2x−y+1=0) and (3, −1) this gives (-17/5, 11/5), like a ball bouncing off a tilted wall. For a triangle, reflect each vertex and connect them-shape is preserved but orientation flips (clockwise ↔ counterclockwise).