A problem on triangle and its perpendicular bisectors.

Question

I'm trying to solve the following problem :

"In △ABC, coordinates of $B$ are $(−3, 3)$. Equation of the perpendicular bisector of side $AB$ is $2x + y − 7 = 0$. Equation of the perpendicular bisector of side $BC$ is $3x − y − 3 = 0$. Mid point of side $AC$ is $E(11/2,7/2)$. Find $AC^2$."

Here is what I did :
By solving $3x − y − 3 = 0$ and $2x + y − 7 = 0$ I find that the intersection of perpendicular bisectors is at $(2,3)$ .
Then using the two points $(2,3)$ and $(11/2,7/2)$, I get the equation of perpendicular bisector of $AC$ as $y = x/7+19/7$.
So the slope of AC is -7 and then using point slope form , $y-7/2=-7(x-11/2)$ Thus the equation of line $AC$ is $y = 42-7 x$ .
Similarly equation of line $BC$ is $y = 2-x/3$ .
So $AC$ and $BC$ intersect at $(6,0)$.
By using the fact that $E$ is the midpoint of $AC$, I find Co-ordinates of A as $(5,7)$.
So the distance between A and C is $5 \sqrt2$, and $AC^2=50$.
But this answer is wrong and the correct answer is $74$ ( I checked the answer sheet) .
What have I done wrong ?

EDIT: Mark Bennet says that $OC^2$ is 74 but clearly they didn't ask for $OC$ since $O$ wasn't in the original given figure. Can anyone please confirm that what I've done is right? Or point out my mistake. Also is there a better way to solve this problem ?

Answer 1

I think your answer is correct. There must be a mis-print in your book.

However, your solution can be shortened by considering the following:-

1. Find O just like you did.
2. O is actually the circum-center of triangle ABC.
3. Then, the circum-radius $= OA = OB = ... = 5$.
4. $OE = ... = \sqrt(50/4)$
5. By Pythagoras theorem, $(AC/2)^2 = 5^2 - (\sqrt(50/4))^2 = ...= 50/4$.