Reflect triangle about 3X-4Y+8=0

Step 1: find the slope and intercept. 3X − 4Y + 8 = 0 gives Y = (3/4)X + 2, so m = 3/4 = 0.75, c = 2, and θ = tan⁻¹(0.75) = 36.87° (cosθ = 0.8, sinθ = 0.6).

Step 2: composite transformation M = T(0,c) · R(θ) · Ref_x · R(−θ) · T(0,−c), which simplifies to

              | 1−m²    2m     −2mc  |
 M = 1/(1+m²) | 2m     m²−1     2c   |
              |  0       0     1+m²  |

With m = 0.75: 1 + m² = 1.5625, 1 − m² = 0.4375, 2m = 1.5, 2mc = 3, 2c = 4. Also cos2θ = 0.28, sin2θ = 0.96.

        | 0.28   0.96  −1.92 |
   M =  | 0.96  −0.28   2.56 |
        |  0      0      1   |

Step 3: apply to the vertices (homogeneous column vectors [x y 1]ᵀ):

A(4,1):  x' = 0.28·4 + 0.96·1 − 1.92 = 1.12 + 0.96 − 1.92 = 0.16
         y' = 0.96·4 − 0.28·1 + 2.56 = 3.84 − 0.28 + 2.56 = 6.12
B(5,2):  x' = 1.40 + 1.92 − 1.92 = 1.40
         y' = 4.80 − 0.56 + 2.56 = 6.80
C(4,3):  x' = 1.12 + 2.88 − 1.92 = 2.08
         y' = 3.84 − 0.84 + 2.56 = 5.56

Check (for A): distance of A from the line = (3·4 − 4·1 + 8)/5 = 3.2; A' = A − 2·3.2·(3,−4)/5 = (4 − 3.84, 1 + 5.12) = (0.16, 6.12). It agrees.

Answer: A'(0.16, 6.12), B'(1.4, 6.8), C'(2.08, 5.56).