The quadrilateral ABCD, where A is (8,-3) and C is (10,11), is a rhombus. Find the relations of coordinates of B and D,if page of rhombus is AB=10

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You should use the information that the diagonal of rhombus are perpendicular and they cut each other in two halves, hence, the slopes of diagonals are negative inverses of each other, such that:

`m_(BD) = -1/(m_AC) => (y_B - y_D)/(x_D - x_B) = (x_C - x_A)/(y_C - y_A) = 2/14 = 1/7 => x_D - x_B = 7(y_B - y_D)`

Since the opposite sides of rhombus are parallel, hence, it slopes are equal, such that:

`m_(AB) = m_(CD)`

`(y_B + 3)/(x_B - 8) = (y_D - 11)/(x_D - 10) `

`(y_D - 11)(x_B - 8) = (y_B + 3)(x_D - 10)`

`m_(BC) = m_(DA)`

`(11 - y_B)/(10 - x_B) = (-3 - y_D)/(8 - x_D)`

`-30 - 10y_D + 3x_B + x_By_D = 88 - 11x_D - 8y_B + x_Dy_B`

Hence, evaluating the relations that you may use to find coordinates of vertices B and D yields `(y_D - 11)(x_B - 8) = (y_B + 3)(x_D - 10), (-3 - y_D)(10 - x_B) = (11 - y_B)(8 - x_D)` and `x_D - x_B = 7(y_B - y_D).`