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QUAD is a quadrilateral with vertices [tex]$Q (-3, 2)$[/tex], [tex]$U (3, 0)$[/tex], [tex]$A (6, -5)$[/tex], and [tex]$D (0, -3)$[/tex]. The slope for [tex]$\overline{QU}$[/tex] is [tex]$\frac{0 - 2}{3 - (-3)} = -\frac{1}{3}$[/tex]. The slope for [tex]$\overline{UA}$[/tex] is [tex]$\frac{-5 - 0}{6 - 3} = -\frac{5}{3}$[/tex]. The slope for [tex]$\overline{AD}$[/tex] is [tex]$\frac{-3 - (-5)}{0 - 6} = -\frac{1}{3}$[/tex]. The slope for [tex]$\overline{DQ}$[/tex] is [tex]$\frac{-3 - 2}{0 - (-3)} = -\frac{5}{3}$[/tex].

Therefore, QUAD is a parallelogram.

What is the missing step in the proof?

A. [tex]$\overline{QU} \parallel \overline{AD}$[/tex] and [tex]$\overline{UA} \parallel \overline{DQ}$[/tex] because the segments have the same slope.
B. [tex]$\overline{QU} \perp \overline{AD}$[/tex] and [tex]$\overline{UA} \perp \overline{DQ}$[/tex] because the segments have the same slope.
C. [tex]$\overline{QU} \parallel \overline{AD}$[/tex] and [tex]$\overline{UA} \parallel \overline{DQ}$[/tex] because the product of the slopes is [tex]$-1$[/tex].
D. [tex]$\overline{QU} \perp \overline{AD}$[/tex] and [tex]$\overline{UA} \perp \overline{DQ}$[/tex] because the product of the slopes is [tex]$-1$[/tex].



Answer :

GinnyAnswer
To determine the missing step in proving that QUAD is a parallelogram, we need to show that the opposite sides are parallel. This can be achieved by confirming that the slopes of opposite sides are equal. Let's analyze the slopes provided:

1. Slope of [tex]\(\overline{QU}\)[/tex]:
[tex]\[ \text{Slope of } \overline{QU} = \frac{2 - 0}{-3 - 3} = \frac{2}{-6} = -\frac{1}{3} \][/tex]

Given slope: [tex]\(-\frac{1}{3}\)[/tex]

2. Slope of [tex]\(\overline{UA}\)[/tex]:
[tex]\[ \text{Slope of } \overline{UA} = \frac{-5 - 0}{6 - 3} = \frac{-5}{3} = -\frac{5}{3} \][/tex]

Given slope: [tex]\(-\frac{5}{3}\)[/tex]

3. Slope of [tex]\(\overline{AD}\)[/tex]:
[tex]\[ \text{Slope of } \overline{AD} = \frac{-3 - (-5)}{0 - 6} = \frac{2}{-6} = -\frac{1}{3} \][/tex]

Given slope: [tex]\(-\frac{1}{3}\)[/tex]

4. Slope of [tex]\(\overline{DQ}\)[/tex]:
[tex]\[ \text{Slope of } \overline{DQ} = \frac{-3 - 2}{0 - (-3)} = \frac{-5}{3} = -\frac{5}{3} \][/tex]

Given slope: [tex]\(-\frac{5}{3}\)[/tex]

Now, let's verify these calculations against the given slopes:
- The slope of [tex]\(\overline{QU}\)[/tex] is [tex]\(-\frac{1}{3}\)[/tex], which agrees with [tex]\(\overline{AD}\)[/tex] having the same slope of [tex]\(-\frac{1}{3}\)[/tex].
- The slope of [tex]\(\overline{UA}\)[/tex] is [tex]\(-\frac{5}{3}\)[/tex], which agrees with [tex]\(\overline{DQ}\)[/tex] having the same slope of [tex]\(-\frac{5}{3}\)[/tex].

Since opposite sides of the quadrilateral QU and AD, and UA and DQ share the same slopes, we conclude that these pairs of sides are parallel.

Thus, the correct missing step in the proof would be:

A. [tex]\(\overline{QU} \| \overline{AD}\)[/tex] and [tex]\(\overline{UA} \| \overline{DQ}\)[/tex] because the segments have the same slope.

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