# Name the quadrilateral formed, if any, by the following points, and give reasons for your answers:$A (-3, 5), B (3, 1), C (0, 3), D (-1, -4)$

Given:

Given points are $A (-3, 5), B (3, 1), C (0, 3), D (-1, -4)$.

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,

$\mathrm{AB}=\sqrt{(3+3)^{2}+(1-5)^{2}}$

Squaring on both sides, we get,

$\mathrm{AB}^{2}=(3+3)^{2}+(1-5)^{2}$

$=(6)^{2}+(-4)^{2}$

$=36+16$

$=52$
$\mathrm{BC}^{2}=(0-3)^{2}+(3-1)^{2}$
$=(-3)^{2}+(2)^{2}$
$=9+4$

$=13$
$\mathrm{CD}^{2}=(-1-0)^{2}+(-4-3)^{2}$

$=(-1)^{2}+(-7)^{2}$
$=1+49$

$=50$
$\mathrm{DA}^{2}=(-3+1)^{2}+(5+4)^{2}$

$=(-2)^{2}+(9)^{2}$
$=4+81$

$=85$
$\mathrm{AC}^{2}=(0+3)^{2}+(3-5)^{2}$

$=(3)^{2}+ (-2)^{2}$
$=9+4$

$=13$
In $\Delta \mathrm{ABC}$,
$\mathrm{AB}=\sqrt{52}, \mathrm{AC}=\sqrt{13}, \mathrm{BC}=\sqrt{13}$
$\mathrm{AC}+\mathrm{BC}=\sqrt{13}+\sqrt{13}=2 \sqrt{13}$
$\mathrm{AB}=\sqrt{52}=2\sqrt{13}$
$\Rightarrow \mathrm{AC}+\mathrm{BC}=\mathrm{AB}$
This implies, the points $A, B ,C$ are collinear.
Therefore, $\Delta \mathrm{ABC}$ is not possible.

Hence, $\mathrm{ABCD}$ is not a quadrilateral.

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Updated on: 10-Oct-2022

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