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 .

1) Coordinates of point C.

C is the midpoint of the segment AB.

So, its coordinates are:

x = (2 + 8) / 2 = 10/2 = 5
y = (3 + 1) / 2 = 4 / 2 = 2

So, the point C is (5,2)

2) Slope of the segment CD.

First, calculate the slope of the segment AB, m

m = Δy / Δx = (y2 - y1) / (x2 - x1) = (3 -1) / (2 - 8) = 2 / ( -6) = - 1/3

Second, calculate the slope of CD,m', as the negative inverse of m:

m' = - 1 / m = - 1 / (-1/3) = 3

Slope of segment CD is 3.

3) Possible coordinates of point D.

You have to use this three data:
Point C: (5, 2)
length CD = √10
slope CD = 3

Call the point D (x,y)

Use fhe equation for the slope: [y - 2] / [x - 5] = 3

=> y - 2 = 3( x- 5)
=> y = 3x - 15 + 2
=> y = 3x - 13

Use the equation for the distance: (x - 5)^2 + (y - 2)^2 = (√10)^2

Expand the parenthesis:
x^2 - 10x + 25 + y^2 - 4y + 4 = 10

Replace y with 3x - 13
x^2 - 10x + 25 + (3x - 13)^2 - 4(3x - 13) + 4 = 10
x^2 - 10x + 25 + 9x^2 - 78x + 169 - 12x + 52 + 4 - 10 = 0

Combine like terms:
10x^2 - 100x + 240 = 0

Simplify by 10:
x^2 - 10x + 24 = 0

Factor
(x - 6)(x - 4) = 0 => x = 6 or x = 4

Calculate y for both x-values:
y = 3(6) - 13 = 5 or y = 3(4) - 13 = - 1

Possible coordinates of point D: (6,5) and (4, -1).