Select the correct answer.

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]\(\overline{AD}\)[/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].

What is the missing step in the proof?

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

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

C. [tex]\( \overline{QU} \perp \overline{AD} \)[/tex] and [tex]\( \overline{UA} \perp \overline{DQ} \)[/tex] because the product of the slopes is -1.

D. [tex]\( \overline{QU} \parallel \overline{AD} \)[/tex] and [tex]\( \overline{UA} \parallel \overline{DQ} \)[/tex] because the product of the slopes is -1.



Answer :

Let's solve the problem step by step.

1. The slope of line segment [tex]\(\overline{QU}\)[/tex]:
Given points are [tex]\(Q(-3,2)\)[/tex] and [tex]\(U(3,0)\)[/tex].

[tex]\[ \text{slope}_{\overline{QU}} = \frac{0 - 2}{3 - (-3)} = \frac{-2}{6} = -\frac{1}{3} \][/tex]

2. The slope of line segment [tex]\(\overline{AD}\)[/tex]:
Given points are [tex]\(A(6,-5)\)[/tex] and [tex]\(D(0,-3)\)[/tex].

[tex]\[ \text{slope}_{\overline{AD}} = \frac{-3 - (-5)}{0 - 6} = \frac{-3 + 5}{0 - 6} = \frac{2}{-6} = -\frac{1}{3} \][/tex]

3. The slope of line segment [tex]\(\overline{DQ}\)[/tex]:
Given points are [tex]\(D(0,-3)\)[/tex] and [tex]\(Q(-3,2)\)[/tex].

[tex]\[ \text{slope}_{\overline{DQ}} = \frac{-3 - 2}{0 - (-3)} = \frac{-5}{3} = -\frac{5}{3} \][/tex]

Now, consider the relationships between the slopes to determine the nature of the lines:

4. Check if lines are parallel by comparing slopes:
- Slope of [tex]\(\overline{QU}\)[/tex] is [tex]\(-\frac{1}{3}\)[/tex].
- Slope of [tex]\(\overline{AD}\)[/tex] is [tex]\(-\frac{1}{3}\)[/tex].
- Slope of [tex]\(\overline{DQ}\)[/tex] is [tex]\(-\frac{5}{3}\)[/tex].

Since [tex]\(\text{slope}_{\overline{QU}} = \text{slope}_{\overline{AD}}\)[/tex], the lines [tex]\(\overline{QU}\)[/tex] and [tex]\(\overline{AD}\)[/tex] are parallel.
- Similarly, as the slopes match [tex]\(\text{slope}_{\overline{UA}}\text{,}\overline{DQ}\)[/tex], these lines are also parallel.

So, the lines with equal slopes are parallel.

5. Check if lines are perpendicular by using the product of slopes:
- The product of the slopes of lines that are perpendicular must be [tex]\(-1\)[/tex].

- Checking product for [tex]\(\overline{QU}\)[/tex] and [tex]\(\overline{AD}\)[/tex] (Should match slopes being perpendicular parallel):
[tex]\[ \text{slope}_{\overline{QU}} \times \text{slope}_{\overline{AD}} = -\frac{5}{3} \times -\frac{1}{3} \implies \frac{5}{9} = 0.555 (greater than 1=-1,meaning lines here should be truly parallel) \][/tex]

- No perpendicular slopes observed meaning the line's relationship is parallel formed not as perpendicular..

Therefore, based on the parallel relationship identified ([tex]\(\overline{QU} \| \overline{AD}\)[/tex] and [tex]\(\overline{DQ}\| \overline{UA})). The correct answer is: B. \(\overline{QU} \parallel \overline{AD}\)[/tex] and [tex]\(\overline{UA} \parallel \overline{DQ}\)[/tex] because the segments have the same slope.