The given equation of a line x – 2y = 3. ∴ Slope of the given line is $\frac{1}{2}$ Let m be the slope of the required line, then tan 45° = $\left|\frac{m - \frac{1}{2}}{1 + m \left(\frac{1}{2}\right)}\right|$ ⇒ 1 = $\left|\frac{2m-1}{2+m}\right|$ ⇒ ± 1 = $\frac{2m-1}{2+m}$ ⇒ 2 + m = 2m – 1 or, – 2 – m = 2m – 1 ⇒ 2 + 1 = 2m – m or, – 2 + 1 = 2m + m ⇒ 3 = m or, – 1 = 3m ⇒ m = 3 or, m = $-\frac{1}{3}$ Case (i) : If m = 3, then the equation of the line passing through (3, 2) is y – 2 = 3(x – 3) ⇒ y – 2 = 3x – 9 ⇒ 3x – y + 2 – 9 = 0 ⇒ 3x – y – 7 = 0. Case (ii) : If m = $-\frac{1}{3}$ , then the equation of the line passing through (3, 2) is y - 2 = $-\frac{1}{3}$ (x - 3) ⇒ 3y – 6 = – x + 3 ⇒ x + 3y – 6 – 3 = 0 ⇒ x + 3y – 9 = 0. ∴ The required equation of lines are 3x – y – 7 = 0 and x + 3y – 9 = 0.