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 square root of 10 units.

A(2, 3)  B(8, 1)
distance AB^2=(8-2)^2+(1-3)^2=40
dAB=2√10
slope mAB=-2/6=-1/3

The equation of the line through the points
y-y1=(-1/3)((x-x1) 
y-3= (-1/3) (x-2) 
y= -(1/3)x +(2/3)+3 = -(1/3)x+11/3 equation of the line AB
perpendicular line has slope 3 (negative reciprocal)
goes through midpoint which is half the distance between the two points
((2+8)/2,(3+1)/2)=(5,2)  Point C
Its equation is y-2=3(x-5) 
y=3x-13 equation of the line CD
calculation of point D    point C (5,2)  point D (x,y)
distance CD=√10
distance CD^2=(x-5)^2+(y-2)^2=10
10=(x-5)^2+(3x-15)^2

solving the quadratic equation
x1=4  y1=-1 
x2=6 y2=5
y=3x-13 equation of the line CD

Point D (6,5)  or (4,-1)

The answer to your question is the last one

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