maths f.3 urgent !!

Threepoints A(4,5) B(-3,2) AND C (x,0) are given .Find the value of x if AC =BC.


Solution 1 (using distance formula) :
AC = BC
AC² = BC²
(4 - x)² + (5 - 0)² = (-3- x)² + (2 - 0)²
16 - 8x + x² + 25 = 9 + 6x + x² +4
-14x = -28
x = 2


Solution 2 (Using slope and mid-point) :

As AC = BC, C lies on the perpendicular bisector of AB.
Let L be the perpendicular bisector of AB :

The coordinates of the mid-point AB
= ((4-3)/2, (5+2)/2)
= (1/2, 7/2)

Slope of AB
= (5 - 2) / (4 + 3)
= 3/7

Slope of L:
[0 - (7/2)] / [x - (1/2)] = -7/3
(0 - 7) / (2x - 1) = -7/3
-7(2x - 1) = -21
2x - 1 = 3
2x = 4
x = 2


However, we CANNOT use ONLY slope or ONLY min-point to find the value of x.

If the x-coordinate of C is 0, it is impossible to have AC = BC...