The equation of a line is x + 3y = 7 ... (i)
⇒ Its slope = −

1

3

If M is the foot of perpendicular from Q on the line (i), then slope of QM = 3.

∴ Equation of QM is y – 8 = 3(x – 3)
⇒ y – 8 = 3x – 9 ⇒ 3x – 9 – y + 8 = 0
⇒ 3x – y – 1 = 0
⇒ 3x – y = 1 ... (ii)
The point M be the foot of perpendicular from Q on the line x + 3y = 7,
which is the point of intersection of lines x + 3y = 7 and 3x – y = 1.
Multiplying (ii) by 3, we get 9x – 3y = 3 ...(iii)
Now adding (i) and (iii), we get 10x = 10 ⇒ x = 1
Put the value of x = 1 in (i), we get 3y = 6 ⇒ y = 2
⇒ The point M is (1, 2)
Let Q′(x′, y′) be the image of Q in the line (i), then M is the mid-point of the segment [Q′Q].
⇒ (1 , 2) = [

3+x′

2

+

8+y′

2

] ⇒

3+x′

2

= 1 ,

8+y′

2

= 2
⇒ 3 + x′ = 2, 8 + y′ = 4 ⇒ x′ = –1, y′ = – 4 ⇒ (x′, y′) = (–1, – 4)
⇒ Q′ is (–1, –4), which is the required image of Q(3, 8) in the line (i).