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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 product of the slopes is [tex]\(-1\)[/tex].

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 segments have the same slope.

Sagot :

GinnyAnsweridn
To determine the missing step in proving that the quadrilateral [tex]\( QUAD \)[/tex] is a parallelogram, we need to verify that both pairs of opposite sides are parallel. Recall that two lines are parallel if they have the same slope.

Let's summarize the given slopes:
1. The slope of [tex]\(\overline{QU}\)[/tex] is [tex]\(-\frac{1}{3}\)[/tex].
2. The slope of [tex]\(\overline{UA}\)[/tex] is [tex]\(-\frac{5}{3}\)[/tex].
3. The slope of [tex]\(\overline{AD}\)[/tex] is [tex]\(-\frac{1}{3}\)[/tex].
4. The slope of [tex]\(\overline{DQ}\)[/tex] is [tex]\(-\frac{5}{3}\)[/tex].

Now we match the slopes of opposite sides to check for parallelism:
- [tex]\(\overline{QU}\)[/tex] and [tex]\(\overline{AD}\)[/tex] both have a slope of [tex]\(-\frac{1}{3}\)[/tex]; hence, these segments are parallel.
- [tex]\(\overline{UA}\)[/tex] and [tex]\(\overline{DQ}\)[/tex] both have a slope of [tex]\(-\frac{5}{3}\)[/tex]; hence, these segments are also parallel.

Therefore, we can conclude:
1. [tex]\(\overline{QU} \parallel \overline{AD}\)[/tex] because they have the same slope.
2. [tex]\(\overline{UA} \parallel \overline{DQ}\)[/tex] because they also have the same slope.

Given this, the correct missing step in the proof is:
A. [tex]\(\overline{QU} \parallel \overline{AD}\)[/tex] and [tex]\(\overline{UA} \parallel \overline{DQ}\)[/tex] because the segments have the same slope.
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