Exercise 1.1.6

Question

If three corners of a parallelogram are (1,1),(4,2) and (1,3), what are all three possible fourth corners? Draw two of them

Neil Z. · Accepted Answer

Let the corners of the parallelogram are: $A=(1,1)$,  $B=(1,3)$, $C=(4,2)$, the fourth corner is $D$. There are three possible values for $D$:
\begin{itemize}
\item If $BC = AD$, then $D = A + AD = A + BC = A + (C-B) = (1,1)+(3,-1) = (4,0)$.
\item If $BD=AC$, then $D = B + BD = B + AC =  B + (C-A) = (4,2) + (3, 1) = (7, 3)$
\item If $AD=CB$, then $D = A + AD = A + CB = A + (B-C) = (1,1) + (-3,1) = (-2, 2)$
\end{itemize}

Arjun Singh Yadav · Answer

It seems above answer is mostly correct except (7,3) is not correct, as it should be:
D=B+(C−A) was computed incorrectly. It should be (1,3)+(3,1)=(4,4)

Exercise 1.1.6

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Exercise 1.1.6

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