Let (a, b) be the required coordinate.

Given that the tangent to the curve y = 2x² – x + 1 is parallel to the line y = 3x + 9.

Slope of the line = 3

∵ the point lies on the curve

⇒ b = 2a² – a + 1 … (1)

Now, y = 2x² – x + 1

Now value of slope at (a, b)

Given that Slope of tangent = Slope of line

⇒ 4a – 1 = 3

⇒ 4a= 4

⇒ a = 1

From (1),

b = 2 – 1 +1

⇒ b = 2