Reflect triangle about y=x+2

The mirror line is y = x + 2, so m = 1 (θ = 45°) and c = 2.

Composite transformation: translate down by 2 (T(0,−2)), rotate by −45°, reflect about the x-axis, rotate by +45°, translate up by 2.

M = T(0,2) · R(45°) · Ref_x · R(−45°) · T(0,−2)
Using the general result with m = 1, c = 2:
              | 1−m²   2m   −2mc |   | 0 2 −4 |   | 0 1 −2 |
 M = 1/(1+m²) | 2m   m²−1    2c  | = | 2 0  4 | / 2 = | 1 0  2 |
              |  0     0   1+m²  |   | 0 0  2 |   | 0 0  1 |

So x' = y − 2 and y' = x + 2.

Apply to the vertices:

A(−1, 0):  x' = 0 − 2 = −2,   y' = −1 + 2 = 1     -> A'(−2, 1)
B(0, −2):  x' = −2 − 2 = −4,  y' = 0 + 2 = 2      -> B'(−4, 2)
C(1, 0):   x' = 0 − 2 = −2,   y' = 1 + 2 = 3      -> C'(−2, 3)

Check: midpoint of A and A' = (−1.5, 0.5) satisfies y = x + 2; AA' has slope (1−0)/(−2+1) = −1, perpendicular to the mirror line.

Answer: A'(−2, 1), B'(−4, 2), C'(−2, 3).