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If the points $A(6, 1)$, $B(p, 2)$, $C(9, 4)$ and $D(7, q)$ are the vertices of a parallelogram $ABCD$, then find the values of $p$ and $q$. Hence, check whether $ABCD$ is a rectangle or not.
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Diagonals of a parallelogram bisect each other. $\therefore$ Co-ordinates of mid point of diagonal $AC$ = Co-ordinates of mid-point of diagonal $BD$. $(\frac{6+9}{2}, \frac{1+4}{2}) = (\frac{p+7}{2}, \frac{2+q}{2})$ (1 mark). $\Rightarrow \frac{p+7}{2} = \frac{15}{2}$ and $\frac{2+q}{2} = \frac{5}{2}$. $\therefore p=8$ and $q=3$ ($\frac{1}{2}$ mark). Diagonal $AC = \sqrt{3^2 + 3^2} = 3\sqrt{2}$ ($\frac{1}{2}$ mark). Diagonal $BD = \sqrt{(-1)^2 + 1^2} = \sqrt{2}$ ($\frac{1}{2}$ mark). $AC \neq BD \therefore ABCD$ is not a rectangle ($\frac{1}{2}$ mark).