Find the equation of the lines through the point (3, 2) which make an angle of 45° with the line x – 2y = 3. 

Let the slope of the required line be m₁.

The given line can be written as y = 1/2x - 3/2, which is of the form y = mx + c.

∴ Slope of the given line = m₂ = 1/2

It is given that the angle between the required line and line x - 2y - 3 =  is 45°.

We know that if θ is the acute angle between line l₁ and l₂ with slopes m₁ and m₂ respectively, then

tanθ = |^{m₁ - m₂}/_{1 + m1m2}|

∴ tan45° = |^{m₁ - m₂}/_{1 + m1m2}|

1 = |^{1/₂ - m₁}/_{1 + ^m1/2}| = |^{1 - 2m₁}/_{2 + m1}|

1 = |^{1 - 2m₁}/_{2 + m1}| ⇒ 1 = ±^{1 - 2m₁}/_{2 + m1}

1 = ^{1 - 2m₁}/_{2 + m1} or 1 = -^{1 - 2m₁}/_{2 + m1}

2 + m₁ = 1 - 2m₁ or 2 + m₁ = -1 + 2m₁

⇒ m₁ = -1/3 or m₁ = 3

Case I: m₁ = 3

The equation of the line passing through (3, 2) and having a slope of 3 is:

y - 2 = 3(x - 3)

y - 2 = 3x - 9

3x - y = 7

Case II: m₁ = -1/3

The equation of the line passing through (3, 2) and having a slope of -1/3 is:

y - 2 = -1/3 (x - 3)

3y - 6 = -x + 3

x + 3y = 9

Therefore, the equations of the lines are 3x - y = 7 and x + 3y = 9.