The image of the point (–8, 12) with respect to the line mirror 4x + 7y + 13 = 0 is

Equation of the given line is4x + 7y + 13 = 0                                                                    (1)Let Q (a, b) be the image of the point P(–8, 12) w.r.t. line (1).Then, PQ ⊥ line (1) and PC = CQ.Equation of the line PC isy-12=74(x+8)[PC is⊥to the line (1) and passes through (–8, 12)]or 7x – 4y + 104 = 0                                                                (2)Solving Eq. (1) and (2), we getx = –12 and y = 5.          ∴C ≡ (–12, 5)Since C is mid-point of PQ,∴-12=α-82 and 5=β+122⇒α=-16 and β=-2∴Q=(-16,-2)