Question

select the correct answer from each drop - down menu.
a quadrilateral has vertices ( a(11,-7), b(9,-4), c(11,-1) ), and ( d(13,-4) ).
rilateral ( a b c d ) is a
if the vertex ( c(11,-1) ) were shifted to the point
quadrilateral ( a b c d ) would be a

Explanation:

Step1: Calculate the lengths of the sides

The distance formula between two points \((x_1,y_1)\) and \((x_2,y_2)\) is \(d = \sqrt{(x_2 - x_1)^2+(y_2 - y_1)^2}\).
For \(AB\): \(x_1 = 11,y_1=-7,x_2 = 9,y_2=-4\), \(d_{AB}=\sqrt{(9 - 11)^2+(-4+7)^2}=\sqrt{(-2)^2+3^2}=\sqrt{4 + 9}=\sqrt{13}\)
For \(BC\): \(x_1 = 9,y_1=-4,x_2 = 11,y_2=-1\), \(d_{BC}=\sqrt{(11 - 9)^2+(-1 + 4)^2}=\sqrt{2^2+3^2}=\sqrt{4+9}=\sqrt{13}\)
For \(CD\): \(x_1 = 11,y_1=-1,x_2 = 13,y_2=-4\), \(d_{CD}=\sqrt{(13 - 11)^2+(-4 + 1)^2}=\sqrt{2^2+(-3)^2}=\sqrt{4 + 9}=\sqrt{13}\)
For \(DA\): \(x_1 = 13,y_1=-4,x_2 = 11,y_2=-7\), \(d_{DA}=\sqrt{(11 - 13)^2+(-7 + 4)^2}=\sqrt{(-2)^2+(-3)^2}=\sqrt{4+9}=\sqrt{13}\)

Step2: Calculate the slopes of the sides

The slope formula between two points \((x_1,y_1)\) and \((x_2,y_2)\) is \(m=\frac{y_2 - y_1}{x_2 - x_1}\)
Slope of \(AB\): \(m_{AB}=\frac{-4+7}{9 - 11}=\frac{3}{-2}=-\frac{3}{2}\)
Slope of \(BC\): \(m_{BC}=\frac{-1 + 4}{11 - 9}=\frac{3}{2}\)
Slope of \(CD\): \(m_{CD}=\frac{-4+1}{13 - 11}=\frac{-3}{2}\)
Slope of \(DA\): \(m_{DA}=\frac{-7 + 4}{11 - 13}=\frac{-3}{-2}=\frac{3}{2}\)

Since \(AB = BC=CD = DA=\sqrt{13}\) and \(m_{AB}\times m_{BC}=-\frac{9}{4}
eq - 1\) (adjacent sides are not perpendicular), the quadrilateral \(ABCD\) is a rhombus with non - perpendicular adjacent sides.

Answer:

rhombus with nonperpendicular adjacent sides