Explanations:
Given the following coordinate points from the kite
[tex]\begin{gathered} W=(a,4b) \\ X=(2a,b) \\ Y=(a,0) \\ Z=(0,b) \end{gathered}[/tex]
For the diagonals to be perpendicular the product of the distance WY and XZ must be zero that is;
[tex]\vec{WY}\cdot\vec{XZ}=0[/tex]
Determine the coordinate point WY
[tex]\begin{gathered} \vec{WY}=[(a-a,4b-0)] \\ \vec{WY}=(0,4b) \end{gathered}[/tex]
Determine the coordinate point XZ
[tex]\begin{gathered} \vec{XZ}=[(2a-0),(b-b)] \\ \vec{XZ}=(2a,0) \end{gathered}[/tex]
Take the dot product of the coordinates
[tex]\begin{gathered} \vec{WY}\cdot\vec{XZ}=(0,4b)\cdot(2a,0) \\ \vec{WY}\cdot\vec{XZ}=[(0)(2a)+(4b)(0))] \\ \vec{WY}\cdot\vec{XZ}=(0,0)=\vec{0} \end{gathered}[/tex]
Since the dot product of the coordinates is a zero vector, hence its diagonals are perpendicular.
