Question

PLEASE HELP

Answer 1

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 $(x_1, y_1)$ and $(x_2, y_2)$ is represented by,

d = $\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Length of WX = $\sqrt{(-1-2)^2+(4-2)^2}$

                       = $\sqrt{9+4}$

                       = $\sqrt{13}$

Length of XY = $\sqrt{(2-0)^2+(2+1)^2}$

                      = $\sqrt{13}$

Length of YZ = $\sqrt{(0+3)^2+(-1-1)^2}$

                      = $\sqrt{13}$

Length of ZW = $\sqrt{(-1+3)^2+(4-1)^2}$

                      = $\sqrt{13}$

Slope of side WX ($m_1$) = $\frac{y_2-y_1}{x_2-x_1}$

                                  = $\frac{4-2}{-1-2}$

                                  = $-\frac{2}{3}$

Slope of side XY ($m_2$) = $\frac{2+1}{2-0}$

                                    = $\frac{3}{2}$

By the property of perpendicular lines,

$m_1\times m_2=-1$

$(-\frac{2}{3})(\frac{3}{2})=-1$

therefore, WX and XY are perpendicular.

Slope of YZ ($m_3$) = $\frac{-1-1}{0+3}=-\frac{2}{3}$

$m_2\times m_3=(\frac{3}{2})\times (-\frac{2}{3})=-1$

Therefore, XY ⊥ YZ

Similarly, we can prove YZ ⊥ ZW.

Therefore, quadrilateral WXYZ is a SQUARE.

Option (1) will be the answer.