Question

Show that the points $(3,-2),(4,0),(6,-3)$ and $(5,-5)$ are the vertices of a parallelogram.

Answer 1

Let $A(3,-2),B(4,0),C(6,-3),D(5,-5)$ be the vertices of a vertices of a parallelogram $ABCD$, when $P(x,y)$ is the mid point of $AC$ & $BD$ respectively.

According to this,

when $P$ is mid point of $AC$

$P(x,y)=(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})$

$=(\frac{3+6}{2},\frac{-2-3}{2})$

$=(\frac{9}{2},\frac{-5}{2})$

when $P$ is midpoint of $BD$

$P(x,y)=(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})$

$=(\frac{4+5}{2},\frac{0-5}{2})$

$=(\frac{9}{2},\frac{-5}{2})$

$\therefore$ The Mid point of $AC$ and $BD$ are equal.

Hence, its a parallelogarm.