Let, coordinate of point A' is (x,y).
Since, A' is the symmetric point A(3, 2) with respect to the line 2x + y - 12 = 0.
So, slope of line containing A and A' will be perpendicular to the line 2x + y - 12 = 0 and also their center lies in the line too.
Now, their center is given by :
Also, product of slope will be -1 .( Since, they are parallel )
x = 2y - 1
So, 
Also, C satisfy given line :
Also,
Therefore, the symmetric points is
.
I'm afraid your equation is not correctly set up. You need to identify the longest side of this right triangle; it is x. This is the "hypotenuse." Next, identify the lengths of the legs: they are sqrt(13) and 2sqrt(2).
Here's a refresher on the Pythagorean Theorem:
(hypotenuse)^2 = (leg 1)^2 + (leg 2)^2
Applying this Theorem here, [x]^2 = [2sqrt(2)]^2 + [sqrt(13)\^2
Solve this for x^2, and then take the positive root (only) of your result.
36/42 in simplest form is 6/7.
The vowels will appear as either EA or AE.
Number of arrangements 2 * 5! = 2 * 120
= 240 answer
