Question

select the correct answer from each drop - down menu.
given: ( w(-1,1) ), ( x(3,4) ), ( y(6,0) ), and ( z(2,-3) ) are the vertices of quadrilateral ( wxyz ).
prove: ( wxyz ) is a square.
using the distance formula, i found that

Explanation:

Step1: Calculate the length of \(WX\)

The distance formula is \(d = \sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\).
For \(W(-1,1)\) and \(X(3,4)\), \(x_1=-1,y_1 = 1,x_2=3,y_2 = 4\).
\(WX=\sqrt{(3-(-1))^2+(4 - 1)^2}=\sqrt{(4)^2+(3)^2}=\sqrt{16 + 9}=\sqrt{25}=5\).

Step2: Calculate the length of \(XY\)

For \(X(3,4)\) and \(Y(6,0)\), \(x_1=3,y_1 = 4,x_2=6,y_2 = 0\).
\(XY=\sqrt{(6 - 3)^2+(0 - 4)^2}=\sqrt{(3)^2+(-4)^2}=\sqrt{9+16}=\sqrt{25}=5\).

Step3: Calculate the length of \(YZ\)

For \(Y(6,0)\) and \(Z(2,-3)\), \(x_1=6,y_1 = 0,x_2=2,y_2=-3\).
\(YZ=\sqrt{(2 - 6)^2+(-3-0)^2}=\sqrt{(-4)^2+(-3)^2}=\sqrt{16 + 9}=\sqrt{25}=5\).

Step4: Calculate the length of \(ZW\)

For \(Z(2,-3)\) and \(W(-1,1)\), \(x_1=2,y_1=-3,x_2=-1,y_2 = 1\).
\(ZW=\sqrt{(-1 - 2)^2+(1-(-3))^2}=\sqrt{(-3)^2+(4)^2}=\sqrt{9 + 16}=\sqrt{25}=5\).

Answer:

all four sides have a length of 5