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…

Diagonals of a parallelogram bisect each other. ∴ Co-ordinates of mid point of diagonal AC = Co-ordinates of mid-point of diagonal BD. \[\begin{aligned} & \left(\frac{6+9}{2}, \frac{1+4}{2}\right)=\left(\frac{\mathrm{p}+7}{2}, \frac{2+\mathrm{q}}{2}\right) \\ & \Rightarrow \frac{\mathrm{p}+7}{2}=\frac{15}{2} \text { and } \frac{2+\mathrm{q}}{2}=\frac{5}{2} \\ & \therefore \mathrm{p}=8 \text { and } \mathrm{q}=3 \end{aligned} \] Diagonal \(\displaystyle \mathrm{AC}=\sqrt{3^{2}+3^{2}}=3 \sqrt{2}\) Diagonal \(\displaystyle \mathrm{BD}=\sqrt{(-1)^{2}+1^{2}}=\sqrt{2}\) \(\displaystyle \mathrm{AC} \neq \mathrm{BD} \therefore \mathrm{ABCD}\) is not a rectangle