Equation of a line passing through the intersection of the lines x - y = 3 and x + y = 0 with inclination 45° is :

1. x + y = 3

2. x - y = 3

3. y - x = 3

4. y = 3x + 1

Given,

Equations :

⇒ x - y = 3 ……..(1)

⇒ x + y = 0 ……..(2)

Adding equation (1) and (2), we get :

⇒ (x - y) + (x + y) = 3 + 0

⇒ x + x - y + y = 3

⇒ 2x = 3

⇒ x = $\dfrac{3}{2}$

Substituting value of x in equation (2), we get :

⇒ $\dfrac{3}{2}$ + y = 0

⇒ y = $-\dfrac{3}{2}$.

∴ Point of intersection of lines x - y = 3 and x + y = 0 is $\Big(\dfrac{3}{2}, -\dfrac{3}{2}\Big)$.

Given,

Inclination (θ) = 45°

Slope = tan 45° = 1.

By point-slope form,

Equation of line : y - y₁ = m(x - x₁)

Equation of line passing through $\Big(\dfrac{3}{2}, -\dfrac{3}{2}\Big)$ and slope = 1 is :

$\Rightarrow y - \Big(-\dfrac{3}{2}\Big) = 1\Big(x - \dfrac{3}{2}\Big) \\[1em] \Rightarrow y + \dfrac{3}{2} = x - \dfrac{3}{2} \\[1em] \Rightarrow x - y = \dfrac{3}{2} + \dfrac{3}{2} \\[1em] \Rightarrow x - y = \dfrac{6}{2} \\[1em] \Rightarrow x - y = 3.$

Hence, Option 2 is the correct option.