The equation of the tangent to the curve y = x² – 2x + 7, which is parallel to the line 2x – y + 9 = 0, is

2x – y + 3 = 0

y = x² – 2x +7

On differentiating with respect to x, we get

$\frac{dy}{dx}=2x-2$ ....(i)

equation of the line 2x – y + 9 = 0

⇒ y = 2x + 9 It's in the form of y = mx + c

So, ∴ Slope of the line = 2 ....(ii)

If a tangent is parallel to the line 2x – y + 9 = 0, then slope of the tangent is equal to slope of the line.

∴ 2 = 2x - 2 → from eq. (i) & (ii)

2 + 2 = 2x

$4 = 2x ⇒ \frac{4}{2}=x=2$

$y=x^2-2x+7$

y = 4 - 4 + 7 = 7

Thus equation of the tangent passing through (2,7) is given by,

$y - y_1 = m (x-x_1)$

$y - 7 = 2 (x - 2)$

$⇒ y-2x-3=0$

Required equation of tangent parallel to line 2x – y + 9 = 0 is

y – 2x – 3 = 0

-2x + y - 3 = 0

2x - y + 3 = 0

Option 1 is correct.