Let A(5, 6), B(2, 8), C(0, 6) & D(3, 4) be the vertices of a parallelogram ABCD, where P(x, y) is the mid-point of AC & BD respectively.
According to this,
When P is the mid-point of AC,
\(P(x, y) = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}2 \right)\)
\(=\left( \frac{5 + 0}{2}, \frac{6 + 6}{2}\right)\)
\(= \left( \frac 52, 6 \right)\)
When P is the mid-point of BD,
\(P(x, y) = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}2 \right)\)
\(=\left( \frac{2+ 3}{2}, \frac{8 + 4}{2}\right)\)
\(= \left( \frac 52, 6 \right)\)
\(\therefore\) The mid-point of AC & BD are equal.
Hence, its a parallelogram.
It is not a square.
