The image of the point (–8, 12) with respect to the line mirror 4x + 7y + 13 = 0 is
(–16, –2)
Equation of the given line is
4x + 7y + 13 = 0 …… (1)
Let Q(α, β) be the image of the point P(–8, 12) with respect to line (1)
Then, PQ ⊥ line (1) and PC = CQ
Equation of the line PC is
(y−12)=74(x+8)
[PC is ⊥ to the line (1) and passes through (–8, 12)]
or 7x – 4y + 104 = 0 …… (2)
solving (1) and (2), we get
x = –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)