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) .
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The coordinates of point C are ((2 + 8)/2, (3 + 1)/2) = (10/2, 4/2) = (5, 2).
The slope of CD is -1/((1 - 3)/(8 - 2)) = -1/(-2/6) = -1/(-1/3) = 3.
Let the cordinates of D be (a, b) then
sqrt((a - 5)^2 + (b - 2)^2) = sqrt(10)
a^2 - 10a + 25 + b^2 - 4b + 4 = 10
a^2 + b^2 - 10a - 4b = -19 . . . (1)
Also, (b - 2)/(a - 5) = 3
b - 2 = 3a - 15
b = 3a - 13 . . . (2)
Putting (2) into (1) gives:
a^2 + (3a - 13)^2 - 10a - 4(3a - 13) = -19
a^2 + 9a^2 - 78a + 169 - 10a - 12a + 52 = -19
10a^2 - 100a + 240 = 0
a^2 - 10a + 24 = 0
(a - 4)(a - 6) = 0
a = 4 or a = 6
When a = 4, b = 3(4) - 13 = 12 - 13 = -1
When a = 6, b = 3(6) - 13 = 18 - 13 = 5

Therefore, the possible coordinates of point D are (6, 5) and (4, -1)

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put like terms together.orremember that the 45 becomes negative when moved to the other side.orthe minuses cancel out.