Question

Prove that the points (-2, -1), (1, 0), (4, 3), and (1, 2) are at the vertices of a parallelogram.

Answer 1

We have the points
$P(-2,-1),Q(1,0),R(4,3)$ and $S(1,2)$
We know the property the of parallelogram that diagonals of parallelogram bisect each other.
Let us find out mid-point of line joining $P$ and $R$ and line joining $Q$ and $S$
(i) Mid-point $M$ of diagonal $PR$
$M(\frac{-2+4}{2},\frac{-1+3}{2})$
$\Rightarrow M(1,1)$
(ii) Mid-point $M^{\prime }$ of diagonal $QS$
$M^{\prime }(\frac{1+1}{2},\frac{0+2}{2})$
$\Rightarrow M^{\prime }(1,1)$
From (i) & (ii)
Mid-points $M$ & $M^{\prime }$ are identical
$\Rightarrow$ Diagonals of the figure $PQRS$ bisect each other and this property is enough to prove that it is a parallelogram.
Although we can also check by distance formula i.e. $d=\sqrt{(a-c{)}^2+(b-d{)}^2}$
$PQ=RS$
$SP=QR$