Answer with explanation:
The four vertices of a two dimensional geometrical shape are set of points A(–1, 2), B(5, 2),C (5, –4), and D(–1, –4).
Now plotting it on X Y plane
Point of intersection of Diagonals that is AC and B D given by the formula
\({\text{Mid point of AC}}=[\frac{x_{1}+x_{2}}{2},\frac{y_{1}+y_{2}}{2}]\\\\=[\frac{-1+5}{2},\frac{-4+2}{2}]\\\\=(2,-1)\\\\{\text{Mid point of BD}}=[\frac{5-1}{2},\frac{2-4}{2}]=(2,-1)\)
Distance formula between two points in two dimensional plane
\(=\sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}\)
\(AB=\sqrt{(5+1)^2+(2-2)^2}=6\\\\BC=6\\\\CD=6, \\\\DA=6\)
So, the Quadrilateral is Rhombus,because all sides are equal.
Slope of line between two points is given by
\(=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)
Slope of AB
\(=\frac{2-2}{5+1}=\frac{0}{6}\)
Slope of BC
\(=\frac{-4-2}{5-5}=\frac{-6}{0}\)
Product of slopes = -1
Hence ∠ABC=90°
If in a rhombus one angle measures 90°,then it a Square.
So, Above Quadrilateral with given vertices is a Square.
