​What are the coordinates of the image of the point (2, 3) in the line (where line acts like a mirror) x + 3y = 5? ​

Correct option is B

​$$\begin{aligned}&\text{Given point } A(2, 3) \text{ and line } x + 3y = 5, \text{ let the image be } B(a, b). \text{ The line acts as a perpendicular} \\&\text{bisector of segment } AB. \\[6p&\text{The slope of } AB \text{ is } \frac{b - 3}{a - 2}, \text{ and the slope of the line is } -\frac{1}{3}. \text{ Since the line is perpendicular to } AB: \\&\frac{b - 3}{a - 2} \cdot \left(-\frac{1}{3}\right) = -1 \Rightarrow \frac{b - 3}{a - 2} = 3 \Rightarrow b - 3 = 3(a - 2) \Rightarrow b - 3 = 3a - 6 \Rightarrow 3a - b = 3 \tag{1} \\[10pt]&\text{The midpoint of } AB \text{ is } \left( \frac{a + 2}{2}, \frac{b + 3}{2} \right), \text{ and this point lies on the line } x + 3y = 5, \text{ so substitute:} \\&\frac{a + 2}{2} + 3 \cdot \frac{b + 3}{2} = 5 \Rightarrow \frac{a + 2 + 3b + 9}{2} = 5 \Rightarrow a + 3b + 11 = 10 \Rightarrow a + 3b = -1 \tag{2} \\[10pt]&\text{Multiply equation (2) by 3:} \quad 3a + 9b = -3 \\[6pt]&\text{Subtract equation (1) from this:} \\&(3a + 9b) - (3a - b) = -3 - 3 \Rightarrow 3a + 9b - 3a + b = -6 \Rightarrow 10b = -6 \Rightarrow b = -\frac{3}{5} \\[10pt]&\text{Substitute } b = -\frac{3}{5} \text{ into equation (1):} \\&3a - (-\frac{3}{5}) = 3 \Rightarrow 3a + \frac{3}{5} = 3 \Rightarrow 3a = 3 - \frac{3}{5} = \frac{12}{5} \Rightarrow a = \frac{4}{5} \\[12pt]&\text{Therefore, the image of point } (2, 3) \text{ in the line } x + 3y = 5 \text{ is:} \\&{\left( \frac{4}{5},\ -\frac{3}{5} \right)}\end{aligned}$$​​