# Name the quadrilateral formed, if any, by the following points, and give reasons for your answers:$A (-1, -2), B (1, 0), C (-1, 2), D (-3, 0)$

Given:

Given points are $A (-1, -2), B (1, 0), C (-1, 2), D (-3, 0)$.

To do:

We have to find the quadrilateral formed, if any, by the given points.

Solution:

We know that,

The distance between two points $\mathrm{A}\left(x_{1}, y_{1}\right)$ and $\mathrm{B}\left(x_{2}, y_{2}\right)$ is $\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}}$.

Therefore,

$\mathrm{AB}=\sqrt{(1+1)^{2}+(0+2)^{2}}$

Squaring on both sides, we get,
$\mathrm{AB}^{2}=(1+1)^{2}+(0+2)^{2}$

$=(2)^{2}+(2)^{2}$
$=4+4$

$=8$
$\mathrm{BC}^{2}=(-1-1)^{2}+(2-0)^{2}$

$=(-2)^{2}+(2)^{2}$
$=4+4$

$=8$
$\mathrm{CD}^{2}=(-3+1)^{2}+(0-2)^{2}$

$=(-2)^{2}+(-2)^{2}$
$=4+4$

$=8$
$\mathrm{DA}^{2}=(-1+3)^{2}+(-2+0)^{2}$

$=(2)^{2}+(-2)^{2}$
$=4+4$

$=8$
$\mathrm{AC}^{2}=(-1+1)^{2}+(2+2)^{2}$
$=(0)^{2}+(4)^{2}$

$=0+16$

$=16$
$\mathrm{BD}^{2}=(-3-1)^{2}+(0-0)^{2}$

$=(-4)^{2}+0$

$=16$

Here,

$AB^2=BC^2=CD^2=DA^2$ and $AC^2=BD^2$

This implies,

$AB=BC=CD=DA$ and $AC=BD$
All the sides are equal and the diagonals are equal to each other.
Therefore, the quadrilateral formed by the points $A, B, C, D$ is a square.

Tutorialspoint

Simply Easy Learning

Updated on: 10-Oct-2022

23 Views