Answer:
The slope of ZW is [tex]\frac{2}{5}[/tex]
The lenght of ZY is [tex]\sqrt{29}[/tex]
Quadrilateral WXYZ is a square
Step-by-step explanation:
Statements
case A) The slope of ZW is [tex]\frac{2}{5}[/tex]
The statement is True
we know that
The formula to calculate the slope between two points is equal to
[tex]m=\frac{y2-y1}{x2-x1}[/tex]
we have
[tex]Z(-3,1),W(2,3)[/tex]
substitute
[tex]m=\frac{3-1}{2+3}[/tex]
[tex]m=\frac{2}{5}[/tex]
case B) The slope of YX is [tex]-\frac{5}{2}[/tex]
The statement is False
we know that
The formula to calculate the slope between two points is equal to
[tex]m=\frac{y2-y1}{x2-x1}[/tex]
we have
[tex]Y(-1,-4),X(4,-2)[/tex]
substitute
[tex]m=\frac{-2+4}{4+1}[/tex]
[tex]m=\frac{2}{5}[/tex]
case C) The lenght of ZY is [tex]\sqrt{29}[/tex]
The statement is True
we know that
the formula to calculate the distance between two points is equal to
[tex]d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}[/tex]
we have
[tex]Z(-3,1),Y(-1,-4)[/tex]
substitute
[tex]d=\sqrt{(-4-1)^{2}+(-1+3)^{2}}[/tex]
[tex]d=\sqrt{(-5)^{2}+(2)^{2}}[/tex]
[tex]d=\sqrt{29}\ units[/tex]
case D) The lenght of WX is [tex]5[/tex]
The statement is False
we know that
the formula to calculate the distance between two points is equal to
[tex]d=\sqrt{(y2-y1)^{2}+(x2-x1)^{2}}[/tex]
we have
[tex]W(2,3),X(4,-2)[/tex]
substitute
[tex]d=\sqrt{(-2-3)^{2}+(4-2)^{2}}[/tex]
[tex]d=\sqrt{(-5)^{2}+(2)^{2}}[/tex]
[tex]d=\sqrt{29}\ units[/tex]
case E) Quadrilateral WXYZ is a square
The statement is True
Because sides ZW and YX are parallel sides (has the same slope)
Slope ZY
we have
[tex]Z(-3,1),Y(-1,-4)[/tex]
[tex]m=\frac{-4-1}{-1+3}[/tex]
[tex]m=-\frac{5}{2}[/tex]
so
ZW and ZY are perpendicular sides (the product of their slopes is equal to minus one)
[tex]-\frac{5}{2}*\frac{2}{5}=-1[/tex]
Slope WX
we have
[tex]W(2,3),X(4,-2)[/tex]
[tex]m=\frac{-2-3}{4-2}[/tex]
[tex]m=-\frac{5}{2}[/tex]
ZY and WX are parallel sides (has the same slope)
distance ZW
we have
[tex]Z(-3,1),W(2,3)[/tex]
[tex]d=\sqrt{(3-1)^{2}+(2+3)^{2}}[/tex]
[tex]d=\sqrt{(2)^{2}+(5)^{2}}[/tex]
[tex]d=\sqrt{29}\ units[/tex]
distance YX
we have
[tex]Y(-1,-4),X(4,-2)[/tex]
[tex]d=\sqrt{(-2+4)^{2}+(4+1)^{2}}[/tex]
[tex]d=\sqrt{(2)^{2}+(5)^{2}}[/tex]
[tex]d=\sqrt{29}\ units[/tex]
The four sides are congruent
case F) [tex]YZ=\sqrt{(-3-(-1))^{2}+(1-4)^{2}}[/tex]
The statement is False
[tex]YZ=\sqrt{(-2)^{2}+(-3)^{2}}[/tex]
[tex]YZ=\sqrt{13}[/tex] ----> is not correct
the value of YZ is [tex]\sqrt{29}[/tex]
