Given equation of line is 

    x +3y = 7 .........1

Let point (a,b) is the image of point (3,8)

Slope of line AB = (b-8)/(a-3)

Since the given line x+3y = 7 is perpendiculart to AB

So (b-8)/(a-3) *(-1/3)  = -1    (Since product of two slopes = -1)

=> (b-8)/(a-3) = 3

=> (b-8) = 3(a-3)

=> b-8 = 3a - 9

=> 3a - b = 9-8

=> 3a - b = 1 .........2

Now mid point of AB = {(a+3)/3, (b+8)/2}

This point also satisfies equation 1

So  (a+3)/2 + 3(b+8)/2 = 7

=> a+3 + 3b + 24 = 2*7

=> a+3b + 27 = 14

=> a+3b = 14-27

=> a+3b = -13 ......3

After solving equation 2 and 3, we get

a = -1, b = -4

So image of the point (3,8) with respest to x+3y = 8 is (-1,-4)