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

2x – 3y = 0 …(i)

4x – 5y = 2 …(ii)

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

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

4x – 6y = 0 …(iii)

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

4x – 5y – 4x + 6y = 2 – 0

⇒ y = 2

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

2x – 3(2) = 0

⇒ 2x – 6 = 0

⇒ 2x = 6

⇒ x = 3

Hence, the point of intersection P(x₁, y₁) is (3, 2)

Now, we know that, when two lines are perpendicular, then the product of their slope is equal to -1

m₁ × m₂ = -1

⇒ Slope of the given line × Slope of the perpendicular line = -1

⇒ The slope of the perpendicular line = 2

So, the slope of a line which is perpendicular to the given line is 2

Then the equation of the line passing through the point (3, 2) having slope 2 is:

y – y₁ = m (x – x₁)

⇒ y – 2 = 2(x – 3)

⇒ y – 2 = 2x – 6

⇒ 2x – y – 6 + 2 = 0

⇒ 2x – y – 4 = 0