Question

consider the graph of quadrilateral wxyz. what is the most specific name for quadrilateral wxyz? rectangle parallelogram rhombus square

Explanation:

Step1: Find the lengths of sides

First, we use the distance formula \(d = \sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\) to find the lengths of the sides.
For \(W(-1,4)\) and \(X(1,1)\):
\(d_{WX}=\sqrt{(1 - (-1))^2+(1 - 4)^2}=\sqrt{(2)^2+(-3)^2}=\sqrt{4 + 9}=\sqrt{13}\)
For \(X(1,1)\) and \(Y(-1,-2)\):
\(d_{XY}=\sqrt{(-1 - 1)^2+(-2 - 1)^2}=\sqrt{(-2)^2+(-3)^2}=\sqrt{4 + 9}=\sqrt{13}\)
For \(Y(-1,-2)\) and \(Z(-3,1)\):
\(d_{YZ}=\sqrt{(-3 - (-1))^2+(1 - (-2))^2}=\sqrt{(-2)^2+(3)^2}=\sqrt{4 + 9}=\sqrt{13}\)
For \(Z(-3,1)\) and \(W(-1,4)\):
\(d_{ZW}=\sqrt{(-1 - (-3))^2+(4 - 1)^2}=\sqrt{(2)^2+(3)^2}=\sqrt{4 + 9}=\sqrt{13}\)
So all sides are equal (\(WX = XY=YZ = ZW=\sqrt{13}\)).

Step2: Check the slopes to see if angles are right angles

Slope of \(WX\): \(m_{WX}=\frac{1 - 4}{1 - (-1)}=\frac{-3}{2}\)
Slope of \(XY\): \(m_{XY}=\frac{-2 - 1}{-1 - 1}=\frac{-3}{-2}=\frac{3}{2}\)
The product of slopes \(m_{WX}\times m_{XY}=\frac{-3}{2}\times\frac{3}{2}=-\frac{9}{4}
eq - 1\), so the angle between \(WX\) and \(XY\) is not a right angle. So it is not a rectangle or square. Since all sides are equal and opposite sides are parallel (we can check slopes of opposite sides: slope of \(WX=\frac{-3}{2}\), slope of \(YZ=\frac{1 - (-2)}{-3 - (-1)}=\frac{3}{-2}=-\frac{3}{2}\); slope of \(XY=\frac{3}{2}\), slope of \(ZW=\frac{4 - 1}{-1 - (-3)}=\frac{3}{2}\), so opposite sides are parallel), and all sides are equal, it is a rhombus.

Answer:

rhombus