Suppose the given two lines intersect at a point P(x₁, y₁). Then, (x₁, y₁) satisfies each of the given equations.

2x – 3y + 1 = 0 …(i)

x + y – 2 = 0 …(ii)

Now, we find the point of intersection of eq. (i) and (ii)

Multiply the eq. (ii) by 2, we get

2x + 2y – 4 = 0 …(iii)

On subtracting eq. (iii) from (i), we get

2x – 3y + 1 – 2x – 2y + 4 = 0

⇒ -5y + 5 = 0

⇒ -5y = -5

⇒ y = 1

Putting the value of y in eq. (ii), we get

x + 1 – 2 = 0

⇒ x – 1 = 0

⇒ x = 1

Hence, the point of intersection P(x₁, y₁) is (1, 1)

The equation of a line parallel to y – axis is of the form

x = a where a is some constant

Given that this equation of the line passing through the point of intersection (1, 1)

Hence, point (1, 1) will satisfy the equation of a line.

Putting x = 1 in the equation y = b, we get

x = a

⇒ 1 = a

or a = 1

Now, required equation of line is x = 1