Question 661700
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Hi,
Point P lies on the perpendicular <u>bisector</u> of segment ST. 
If S has coordinates (5,14), T has coordinates (-3,2)
Midpoint segment ST
(5,14), 
(-3,2) Midpoint = (1,8) and slope m = (14-2)/(5-(-3)) = 12/8 = 3/2

Point P(-a/2,a) <u>lies on</u> the perpendicular bisector: slope -2/3 
(-a/2,a)
 (1, 8) slope m = {{{(a-8)/((-a/2)-1) = -2(a-8)/(a+2) = -2/3}}}
| Cross Multiplying to solve
 -6(a-8) = -2(a+2)
 -6a + 48 = -2a - 4
        52 = 4a
        13 = a
and...
(-13/2,13)
(  1 , 8) {{{ m = 5/(-15/2) =  - 10/15 = -2/3}}}