Question

The point $(3,-2),(4,0),(6,-3)$ and $(5,-5)$ are the vertices of a parallelogram.

• A. True
• B. False

Answer 1

The correct option is A True

Let $P(x,y)$ be the point of intersection of diagonals $AC$ and $BD$ of $ABCD.$

Mid- point of $AC:$

$x=\frac{3+6}{2}=\frac{9}{2}$

$y=\frac{-2-3}{2}=\frac{-5}{2}$

Mid point of $AC=(\frac{9}{5},\frac{-5}{2})$

Now, mid-point of $BD:$

$x=\frac{5+4}{2}=\frac{9}{2}$

$y=\frac{-5+0}{2}=\frac{-5}{2}$

Mid-point of $BD=(\frac{9}{2},-\frac{5}{2})$

Here, mid-point of $AC$ and mid-point of $BD$ i.e, diagonals $AC$ and $BD$ bisect each other.

We know that diagonals of a parallelogram bisect each other.

$\therefore$ $ABCD$ is a parallelogram.