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Given the position vectors of three 2-dimensional points $A$, $B$ and $C$, find the coordinates of a fourth point $D$ such that $ABCD$ forms a parallelogram. 

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There are 3 points $A$, $B$, $C$, with position vectors \\[ \\mathbf a = \\var{a},\\quad \\mathbf b = \\var{b},\\quad \\mathbf c= \\var{c} \\]

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respectively.

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Find the coordinates of the point $D$ such that $ABCD$ forms a parallelogram.

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We want to find the position vector $\\mathbf d = \\vec{OD}$, such that $ABCD$ forms a parallelogram.

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{geogebra_applet('https://www.geogebra.org/m/jqaptsj6',defs)}

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For $ABCD$ to be a parallelogram, then $\\vec{AB} = \\vec{DC}$. In terms of the points' position vectors,

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\\[ \\begin{split} \\vec{AB} &\\,= \\mathbf b - \\mathbf a &\\,= \\var{b-a} ,\\\\\\\\ \\vec{DC} &\\,= \\mathbf c - \\mathbf d &\\,= \\var{c} - \\mathbf d. \\end{split} \\]

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Therefore, \\[ \\begin{split} \\var{b-a} &\\,= \\var{c} -\\mathbf d \\\\ \\implies \\mathbf d &\\,= \\var{c+a-b}. \\end{split}\\]

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Hence, the co-ordinates of $D$ are $(\\var{d[0]},\\var{d[1]})$.

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$D =$ [[0]]

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