The given equation of a line x – 2y = 3.
∴ Slope of the given line is

1

2

Let m be the slope of the required line, then tan 45° = |

m− 1 2

1+m( 1 2 )

|
⇒ 1 = |

2m−1

2+m

| ⇒ ± 1 =

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 = −

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 = −

1

3

, then the equation of the line passing through (3, 2) is
y - 2 = −

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.