A(-1,-1), B(6,1), C(8,8), D(x,y) are the four vertices of a rhombus taken in order. Find the co-ordinates of point D.

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Solution

$\begin{matrix} A B = \sqrt{(6 + D^{2}) + (1 + D^{2})} & A D = \sqrt{(x + 1)^{2} (y + 1)^{2}} = \sqrt{4 a + 4} & = \sqrt{x^{2} + 2 x + 1 + y^{2} + 2 y + 1} = \sqrt{51} & = \sqrt{x^{2} + y^{2} + 2 x + 2 y + 2} \end{matrix}$
$\begin{matrix} D C = \sqrt{(x + 8)^{2} + (y + 8)^{2}} & B C = \sqrt{(8 - 6)^{2} (8 - 1)^{2}} = \sqrt{x^{2} - 16 x + 64 + y^{2} - 16 y + 64} & = \sqrt{4 + 49} = \sqrt{x^{2} + y^{2} - 16 x - 16 y + 128} & = \sqrt{53} \end{matrix}$
AB =AD & BC =DC

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A(-1,-1), B(6,1), C(8,8), D(x,y) are the four vertices of a rhombus taken in order. Find the co-ordinates of point D.

AB=√(6+D2)+(1+D2)AD=√(x+1)2(y+1)2=√4a+4=√x2+2x+1+y2+2y+1=√51=√x2+y2+2x+2y+2 DC=√(x+8)2+(y+8)2BC=√(8−6)2(8−1)2=√x2−16x+64+y2−16y+64=√4+49=√x2+y2−16x−16y+128=√53 AB =AD & BC =DC