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ABCD is a rectangle formed by the points $A (-1, -1)$, $B (-1, 6)$, $C (3, 6)$ and $D (3, -1)$. P, Q, R and S are mid-points of sides AB, BC, CD and DA respectively. Show that diagonals of the quadrilateral PQRS bisect each other.
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Sol. Co-ordinates of point P are $\left(\frac{-1-1}{2}, \frac{-1+6}{2}\right)$ i.e. $\left(-1,\frac{5}{2}\right)$
Co-ordinates of point Q are $\left(\frac{-1+3}{2}, \frac{6+6}{2}\right)$ i.e. $(1, 6)$
Co-ordinates of point R are $\left(\frac{3+3}{2}, \frac{6-1}{2}\right)$ i.e. $\left(3,\frac{5}{2}\right)$
Co-ordinates of point S are $\left(\frac{-1+3}{2}, \frac{-1-1}{2}\right)$ i.e. $(1,-1)$
Co-ordinates of mid point of diagonal QS are $\left(\frac{1+1}{2}, \frac{6-1}{2}\right)$ i.e. $\left(1,\frac{5}{2}\right)$
Co-ordinates of mid point of diagonal PR are $\left(\frac{-1+3}{2}, \frac{5+5}{2}\right)$ i.e. $\left(1,\frac{5}{2}\right)$
Since coordinates of mid point of QS = coordinates of mid point of PR
Therefore, diagonals PR and QS bisect each other.
Co-ordinates of point Q are $\left(\frac{-1+3}{2}, \frac{6+6}{2}\right)$ i.e. $(1, 6)$
Co-ordinates of point R are $\left(\frac{3+3}{2}, \frac{6-1}{2}\right)$ i.e. $\left(3,\frac{5}{2}\right)$
Co-ordinates of point S are $\left(\frac{-1+3}{2}, \frac{-1-1}{2}\right)$ i.e. $(1,-1)$
Co-ordinates of mid point of diagonal QS are $\left(\frac{1+1}{2}, \frac{6-1}{2}\right)$ i.e. $\left(1,\frac{5}{2}\right)$
Co-ordinates of mid point of diagonal PR are $\left(\frac{-1+3}{2}, \frac{5+5}{2}\right)$ i.e. $\left(1,\frac{5}{2}\right)$
Since coordinates of mid point of QS = coordinates of mid point of PR
Therefore, diagonals PR and QS bisect each other.