Question

quadrilateral abcd has the following vertices:

• a(-6, -2)
• b(-4, 4)
• c(8, 1)
• d(6, -6)

also, ∠a is a right angle.
is quadrilateral abcd a rectangle, and why?
choose 1 answer:
a yes, because ab = bc = cd = ad, and ∠a is a right angle.
b yes, because opposite sides are parallel, and ∠a is a right angle.
c no, because \\(\overline{ab}\\) is not parallel to \\(\overline{cd}\\).
d no, because \\(\overline{ab}\\) is shorter than \\(\overline{bc}\\).

Explanation:

Step1: Recall Rectangle Properties

A rectangle is a quadrilateral with four right angles and opposite sides parallel and equal. Also, if a quadrilateral has a right angle and opposite sides parallel, it can be a rectangle. We need to check the slopes of sides to determine parallelism. The slope formula is $m = \frac{y_2 - y_1}{x_2 - x_1}$.

Step2: Calculate Slope of AB

For points $A(-6, -2)$ and $B(-4, 4)$, slope of $AB$: $m_{AB}=\frac{4 - (-2)}{-4 - (-6)}=\frac{6}{2} = 3$.

Step3: Calculate Slope of CD

For points $C(8, 1)$ and $D(6, -6)$, slope of $CD$: $m_{CD}=\frac{-6 - 1}{6 - 8}=\frac{-7}{-2}=\frac{7}{2}$. Wait, no, wait, let's recalculate. Wait, $C(8,1)$ and $D(6, -6)$: $y_2 - y_1=-6 - 1=-7$, $x_2 - x_1=6 - 8=-2$, so $m_{CD}=\frac{-7}{-2}=\frac{7}{2}$? Wait, no, maybe I made a mistake. Wait, let's check slope of AD and BC too.

Wait, slope of AD: $A(-6, -2)$ and $D(6, -6)$: $m_{AD}=\frac{-6 - (-2)}{6 - (-6)}=\frac{-4}{12}=-\frac{1}{3}$.

Slope of BC: $B(-4, 4)$ and $C(8, 1)$: $m_{BC}=\frac{1 - 4}{8 - (-4)}=\frac{-3}{12}=-\frac{1}{4}$. Wait, no, that can't be. Wait, no, maybe I messed up. Wait, the problem says angle A is a right angle, so AB and AD should be perpendicular. Let's check slopes of AB and AD. Slope of AB is 3, slope of AD is $-\frac{1}{3}$. The product of slopes of perpendicular lines is -1. $3\times(-\frac{1}{3})=-1$, so AB and AD are perpendicular (right angle at A). Now check opposite sides: AB and CD, AD and BC.

Wait, slope of AB is 3, slope of CD: let's recalculate CD. $C(8,1)$, $D(6, -6)$: $y_2 - y_1=-6 - 1=-7$, $x_2 - x_1=6 - 8=-2$, so $m_{CD}=\frac{-7}{-2}=\frac{7}{2}$? Wait, no, that's not 3. Wait, maybe I made a mistake. Wait, no, let's check slope of AB again: $A(-6,-2)$, $B(-4,4)$: $4 - (-2)=6$, $-4 - (-6)=2$, so slope 3. Correct. Slope of CD: $D(6,-6)$, $C(8,1)$: $1 - (-6)=7$, $8 - 6=2$, so slope $\frac{7}{2}$? Wait, no, $C$ to $D$: $y$ changes from 1 to -6 (decrease by 7), $x$ changes from 8 to 6 (decrease by 2), so slope is $\frac{-7}{-2}=\frac{7}{2}$. So AB slope is 3, CD slope is $\frac{7}{2}$: not equal, so AB is not parallel to CD? Wait, but option C says "No, because $\overline{AB}$ is not parallel to $\overline{CD}$". Wait, but let's check slope of AD and BC. Slope of AD: $A(-6,-2)$, $D(6,-6)$: $-6 - (-2)=-4$, $6 - (-6)=12$, slope $-\frac{1}{3}$. Slope of BC: $B(-4,4)$, $C(8,1)$: $1 - 4=-3$, $8 - (-4)=12$, slope $-\frac{1}{4}$. Wait, that's not equal. Wait, but the problem says angle A is a right angle. Wait, maybe I miscalculated slopes. Wait, let's recalculate slope of CD: $C(8,1)$, $D(6,-6)$: $y_2 - y_1=-6 - 1=-7$, $x_2 - x_1=6 - 8=-2$, so slope is $\frac{-7}{-2}=\frac{7}{2}$. Slope of AB is 3. 3 is not equal to $\frac{7}{2}$, so AB is not parallel to CD. So option C says "No, because $\overline{AB}$ is not parallel to $\overline{CD}$", which matches. Wait, but let's check the options again. Option A: says AB=BC=CD=AD, which would be a square or rhombus, but let's check lengths. Length of AB: $\sqrt{(-4 + 6)^2 + (4 + 2)^2}=\sqrt{4 + 36}=\sqrt{40}$. Length of BC: $\sqrt{(8 + 4)^2 + (1 - 4)^2}=\sqrt{144 + 9}=\sqrt{153}$. Not equal, so A is wrong. Option B: says opposite sides are parallel, but we saw AB and CD are not parallel, so B is wrong. Option C: says AB not parallel to CD, which is true. Option D: says AB is shorter than BC, AB length is $\sqrt{40}\approx6.32$, BC length is $\sqrt{153}\approx12.37$, so AB is shorter, but that's not the reason for not being a rectangle. The key is parallelism. So the correct answer is C.

Answer:

C. No, because $\overline{AB}$ is not parallel to $\overline{CD}$.