Answer is C
Let (x,y) be the image of given point (3,8). Line joining point (3, 8) and image (x, y) is perpendicular to given line x + 3y = 7
So, $m_{1}*m_{2} =-1$ -----(1)
For the line x + 3y = 7, 3y = -x + 7 → y = $\frac{-x+7}{3}$, so [ y = mx + c; where m is the slope]
slope $m_{1}=\frac{-1}{3}$ ------(2)
Slope of line joining the point (3, 8) and the image (x, y) is $m_{2}$ =$\frac{y_{2}-y_{1}}{x_{2}-x_{1}} = \frac{y-8}{x-3}$ ------(3)
Putting (2) and (3) in (1), we will get 3x – y – 1 = 0
On solving both equations of lines x + 3y = 7 and 3x – y = -1, we will get the midpoint (x,y) = (1, 2)
In order to find the image (x, y), $\frac{3+x}{2}=1; \frac{8+y}{2}=2;$
On solving, we will get image as (-1, -4)