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). quadrilateral abcd is a drop - down menu. if the vertex c(11, -1) were shifted to the point c(11, 1), quadrilateral abcd drop - down menu. the options for the first drop - down menu are: parallelogram with nonperpendicular and noncongruent adjacent sides, trapezoid with exactly one pair of parallel sides, rectangle with noncongruent adjacent sides, rhombus with nonperpendicular adjacent sides

Explanation:

Step1: Calculate side lengths of ABCD

Use distance formula $d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$:

• $AB=\sqrt{(9-11)^2+(-4+7)^2}=\sqrt{4+9}=\sqrt{13}$
• $BC=\sqrt{(11-9)^2+(-1+4)^2}=\sqrt{4+9}=\sqrt{13}$
• $CD=\sqrt{(13-11)^2+(-4+1)^2}=\sqrt{4+9}=\sqrt{13}$
• $DA=\sqrt{(11-13)^2+(-7+4)^2}=\sqrt{4+9}=\sqrt{13}$

Step2: Check side perpendicularity

Use slope formula $m=\frac{y_2-y_1}{x_2-x_1}$:

• Slope of $AB$: $m_{AB}=\frac{-4+7}{9-11}=\frac{3}{-2}=-\frac{3}{2}$
• Slope of $BC$: $m_{BC}=\frac{-1+4}{11-9}=\frac{3}{2}$

Product: $m_{AB} \times m_{BC}=-\frac{3}{2} \times \frac{3}{2}=-\frac{9}{4}
eq -1$, so sides not perpendicular.

Step3: Analyze shifted quadrilateral ABC'D

New vertex $C'(11,1)$:

• $BC'=\sqrt{(11-9)^2+(1+4)^2}=\sqrt{4+25}=\sqrt{29}$
• $C'D=\sqrt{(13-11)^2+(-4-1)^2}=\sqrt{4+25}=\sqrt{29}$
• $AB=DA=\sqrt{13}$, $BC'=C'D=\sqrt{13}$
• Slopes of $AB$ and $C'D$: $m_{C'D}=\frac{-4-1}{13-11}=-\frac{5}{2}$; $m_{AB}=-\frac{3}{2}$ (not parallel)
• Slopes of $BC'$ and $DA$: $m_{BC'}=\frac{1+4}{11-9}=\frac{5}{2}$; $m_{DA}=\frac{-7+4}{11-13}=\frac{3}{2}$ (not parallel)
• Slopes of $AB$ and $BC'$: $m_{AB} \times m_{BC'}=-\frac{3}{2} \times \frac{5}{2}=-\frac{15}{4}

eq -1$ (not perpendicular)

• Only $AD \parallel BC$ (original) and $AB \parallel DC$ (original) no longer hold; but original ABCD has all sides equal, non-perpendicular adjacent sides.

Answer:

rhombus with nonperpendicular adjacent sides