Given:

The curve y = x² – 4x + 5 and line is 2y + x = 7

y = x² – 4x + 5

Differentiating the above w.r.t x,

we get the Slope of the tangent,

⇒ = 2x^{2 – 1} – 4 + 0

⇒ = 2x – 4 ...(1)

Since, line is 2y + x = 7

⇒ 2y = – x + 7

⇒ y =

⇒ y = +

The equation of a straight line is y = mx + c, where m is the The Slope of the line.

Thus, the The Slope of the line is ...(2)

Since, tangent is perpendicular to the line,

The Slope of the normal =

From (1) & (2),we get

i.e, =

⇒ 1 =

⇒ x – 2 = 1

⇒ x = 3

Substituting x = 3 in y = x² – 4x + 5,

⇒ y = y = 3² – 4×3 + 5

⇒ y = 9 – 12 + 5

⇒ y = 2

Thus, the required point is (3,2)