Answer:
Option (1)
Step-by-step explanation:
Coordinates of the vertices of a quadrilateral WXYZ drawn in the figure are,
W(-1, 4), X(2, 2), Y(0, -1), Z(-3, 1)
Length of a segment having ends as [tex](x_1, y_1)[/tex] and [tex](x_2, y_2)[/tex] is represented by,
d = [tex]\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}[/tex]
Length of WX = [tex]\sqrt{(-1-2)^2+(4-2)^2}[/tex]
= [tex]\sqrt{9+4}[/tex]
= [tex]\sqrt{13}[/tex]
Length of XY = [tex]\sqrt{(2-0)^2+(2+1)^2}[/tex]
= [tex]\sqrt{13}[/tex]
Length of YZ = [tex]\sqrt{(0+3)^2+(-1-1)^2}[/tex]
= [tex]\sqrt{13}[/tex]
Length of ZW = [tex]\sqrt{(-1+3)^2+(4-1)^2}[/tex]
= [tex]\sqrt{13}[/tex]
Slope of side WX ([tex]m_1[/tex]) = [tex]\frac{y_2-y_1}{x_2-x_1}[/tex]
= [tex]\frac{4-2}{-1-2}[/tex]
= [tex]-\frac{2}{3}[/tex]
Slope of side XY ([tex]m_2[/tex]) = [tex]\frac{2+1}{2-0}[/tex]
= [tex]\frac{3}{2}[/tex]
By the property of perpendicular lines,
[tex]m_1\times m_2=-1[/tex]
[tex](-\frac{2}{3})(\frac{3}{2})=-1[/tex]
therefore, WX and XY are perpendicular.
Slope of YZ ([tex]m_3[/tex]) = [tex]\frac{-1-1}{0+3}=-\frac{2}{3}[/tex]
[tex]m_2\times m_3=(\frac{3}{2})\times (-\frac{2}{3})=-1[/tex]
Therefore, XY ⊥ YZ
Similarly, we can prove YZ ⊥ ZW.
Therefore, quadrilateral WXYZ is a SQUARE.
Option (1) will be the answer.
