Find an equation of the normal line to the parabola y = x2 â 9x + 7 that is parallel to the line x â 7y = 2.

Please, could you proof read your input before posting a question?    The character  " â " does not belong here.  "x2" should be written as x^2, where the "^" denotes exponentiation.

I'm taking the liberty of correcting your input:  "Find an equation of the normal line to the parabola y = x2 â 9x + 7 that is parallel to the line x â 7y = 2." may be:

Find an equation of the normal line to the parabola y = x^2 -  9x + 7 that is parallel to the line x - 7y = 2.  

How do you find the slope of a normal line to the graph of a given quadratic?  Differentiate    y = x^2 - 9x + 7:     dy/dx = 2x - 9

You can see that this slope varies with x.  You have not named an x-value here, so   I will have to stop   at   dy/dx = 2x - 9.

The slope of a line perpendicular to the tangent line is the negative reciprocal of   the slope of the tangent line, or

-1
  which obviously depends upon the x value you choose or are given.
(2x)

Now focus on the given line x - 7y = 2.  Solving for y, 7y =x-2.  The slope of this line is (1/7).  Thus, 

    -1            1
=     Cross-multiplying, -7 = 2x; x = -3.5
    2x            7

So it appears we DO have the x-value at the point of tangency on the curve.

Find the associated y-value by subst. x = -3.5 into the eq'n of the parabola.

Call it Y.

Then the eq'n of the normal line to the parabola that is parallel to x - 7y = 2

is             y-Y = (1/7)(x-[-3.5])