Answer:
Option (4). Rhombus
Step-by-step explanation:
From the figure attached,
Distance AB = [tex]\sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}[/tex]
= [tex]\sqrt{(1-4)^2+(-5+3)^2}[/tex]
= [tex]\sqrt{(-3)^2+(-2)^2}[/tex]
= [tex]\sqrt{13}[/tex]
Distance BC = [tex]\sqrt{(4-1)^2+(-3+1)^2}[/tex]
= [tex]\sqrt{9+4}[/tex]
= [tex]\sqrt{13}[/tex]
Distance CD = [tex]\sqrt{(-2-1)^2+(-3+1)^2}[/tex]
= [tex]\sqrt{9+4}[/tex]
= [tex]\sqrt{13}[/tex]
Distance AD = [tex]\sqrt{(1+2)^2+(-5+3)^2}[/tex]
= [tex]\sqrt{9+4}[/tex]
= [tex]\sqrt{13}[/tex]
Slope of AB ([tex]m_{1}[/tex]) = [tex]\frac{y_{2}-y_{1}}{x_{2}-x_{1}}[/tex]
= [tex]\frac{4-1}{-3+5}[/tex]
= [tex]\frac{3}{2}[/tex]
Slope of BC ([tex]m_{2}[/tex]) = [tex]\frac{4-1}{-3+1}[/tex]
= [tex]-\frac{3}{2}[/tex]
If AB and BC are perpendicular then,
[tex]m_{1}\times m_{2}=-1[/tex]
But it's not true.
[[tex]m_{1}\times m_{2}=(\frac{3}{2})(-\frac{3}{2})[/tex] = -[tex]\frac{9}{4}[/tex]]
It shows that the consecutive sides of the quadrilateral are not perpendicular.
Therefore, ABCD is neither square nor a rectangle.
Slope of diagonal BD = [tex]\frac{4+2}{-3+3}[/tex]
= Not defined (parallel to y-axis)
Slope of diagonal AC = [tex]\frac{1-1}{-1+5}[/tex]
= 0 [parallel to x-axis]
Therefore, both the diagonals AC and BD will be perpendicular.
And the quadrilateral formed by the given points will be a rhombus.
