Since we know that both the rays reflected and incident are equally inclined to the normal at A. Now, coordinates of A = (x, 0). If AR makes an angle q with the normal at A, then it forms an angle 90° + θ with the positive x-axis and AS forms an angle 90° – θ with the positive x-axis. Now, slope of AR = tan(90° + θ) = – cotθ and slope of AS = tan(90° – θ) = cotθ ∴ Slope of AS + slope of AR = 0 ⇒ $\frac{3-0}{5-1x} + \frac{0-2}{x-1}$ = 0 ⇒ $\frac{3}{5-x} + \frac{-2}{x-1}$ = 0 ⇒ 3(x – 1) + (–2) (5 – x) = 0 ⇒ 3x – 3 – 10 + 2x = 0 ⇒ 5x – 13 = 0 ⇒ x = $\frac{13}{5}$ ∴ The co-ordinate of A is $\left(\frac{13}{5},0\right)$.