The points A(4, 3), B(6, 4), C(5, – 6)


The points A(4, 3), B(6, 4), C(5, – 6) and D(- 3, 5) are vertices of a parallelogram.



Now, distance between $A(4,3)$ and $B(6,4), A B=\sqrt{(6-4)^{2}+(4-3)^{2}}=\sqrt{2^{2}+1^{2}}=\sqrt{5}$

$\left[\because\right.$ distance between the points $\left(x_{1}, y_{1}\right)$ and $\left.\left(x_{2}, y_{2}\right), d=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}\right]$

Distance between $B(6,4)$ and $C(5,-6), B C=\sqrt{(5-6)^{2}+(-6-4)^{2}}$



Distance between $C(5,-6)$ and $D(-3,5), C D=\sqrt{(-3-5)^{2}+(5+6)^{2}}$



Distance between $D(-3,5)$ and $A(4,3), D A=\sqrt{(4+3)^{2}+(3-5)^{2}}$



In parallelogram, opposite sides are equal. Here, we see that all sides AB, BC, CD and DA are different.

Hence, given vertices are not the vertices of a parallelogram.

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