The endpoints of AB are A(2, 3) and B(8, 1). The perpendicular bisector of AB is CD, and point C lies on AB. The length of CD is (root of 10) units.The coordinates of point C are (-6, 2) (5, 2) (6, -2) (10, 4) . The slope of is -3 -1/3 1/3 3 . The possible coordinates of point D are (4, 5) (5, 5) (6, 5) (8, 3) and (2, 1) (4, -1) (5, -1) (6, -1) .Please pick 1 answer for all the questions

Comment
If C lies on AB Then it must be the midpoint of AB because CD is the Perpendicular Bisector of AB

Midpoint
A(2,3) B(8,1)
Midpoint formula = (x1 + x2)/2 + (y1 + y2)/2
Midpoint = (2 + 8)/2 + (3 + 1) / 2 
Midpoint = 10/2 + 4/2
Midpoint = (5,2) <<<<< answer


Slope AB
slope = (y2 - y1)/(x2 - x1)
slope = (3 - 1) / (2 - 8)
slope = 2/-6 = -1/3

Slope CD
CD_slope * AB_slope = - 1
CD_slope * -1/3 = -1 multiply both sides by - 3
CD_slope * 1 = - 1 * -3
CD_slope = 3   <<<<< Slope CD Answer

Equation CD
Given the midpoint of AB which is C( and the slope of CD
y = 3x + b
2 = 3*5 + b
b = - 13
Equation y = 3x - 13

Using Distance to get D
d^2 = (x1 - x2)^2 + (y2 - y1)'^2 
d^2 = (x2 - 5)^2 + (y2 - 2)^2
y2 = 3*x2 - 13
10 = x2^2 - 10x2 + 25 + (3x2 - 13 - 2)^2
10 = x2^2 - 10x2 + 25 + (3x2 - 15)^2
10 = x1^2 - 10x2 + 25 + 9x^2 - 90x2 + 225
10 = 10x2^2 - 100x2 +250
0 = 10x2^2 - 100x2 + 240 Divide through by 10
0 = x2^2 - 10x^2 + 24
0 = (x2 - 6)(x2 - 4)
 
x2 = 6 or
x2 = 4

Using y2 = 3x - 13
for x2  = 4
then y2 = 3(4) - 13
y2 = - 1

for x2 = 6
y2 = 3*6 - 13
y2 = 5

Possible Answers  for D
D = (4,-1)
D = (6,5)

I've included a graph to show a graphical solution. 

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