The coordinates of AB are A(2,3) and B(8,1) The perpendicular bisected of AB is CD and point C lies on AB the length of CD is 10 units

The Coordinates of Point C are C(5,2)

The slope of CD is 3

The possible coordinates of point D(a,b) are (6, 5) and (4, -1)

Step-by-step explanation:

According to the Question,

Now, Let The Coordinates of Point C are C( x , y ) .

        The Coordinates of Point C are C( x , y ) =   ⇒C(5, 2).

       Thus, The slope of CD is  =    ⇔  3.

       ⇒ √{ (b - 2)² +(a - 5)² } = √10

      on squaring both sides we get,

       ⇒ a² - 10a + 25 + b² - 4b + 4 = 10

       ⇒ a² + b² - 10a - 4b = - 19 ⇔⇔ (Equation 1)

     ⇒Thus,  (b - 2)/(a - 5) = 3

     ⇒ b - 2 = 3a - 15

     ⇒ b = 3a - 13 ⇔⇔ (Equation 2)

 ⇒a² + (3a - 13)² - 10a - 4(3a - 13) = - 19

 ⇒a² + 9a² - 78a + 169 - 10a - 12a + 52 = - 19

 ⇒10a² - 100a + 240 = 0

 ⇒a² - 10a + 24 = 0

 ⇒(a - 4)(a - 6) = 0

 ⇒a = 4 or a = 6

Now,

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

this doesn't make sense lol

this may not be at all, but i use the same website you're using for school, and if i'm stuck doing one of these i try to look for it on the section.