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Quadrilateral [tex][tex]$WXYZ$[/tex][/tex] is on a coordinate plane. Segment [tex][tex]$XY$[/tex][/tex] is on the line [tex][tex]$x-3y=-12$[/tex][/tex], and segment [tex][tex]$WZ$[/tex][/tex] is on the line [tex][tex]$x-3y=-6$[/tex][/tex]. Which statement proves how segments [tex][tex]$XY$[/tex][/tex] and [tex][tex]$WZ$[/tex][/tex] are related?

A. They have opposite reciprocal slopes of [tex][tex]$\frac{1}{3}$[/tex][/tex] and -3 and are, therefore, perpendicular.
B. They have the same slope of [tex][tex]$-\frac{1}{3}$[/tex][/tex] and are, therefore, parallel.
C. They have opposite reciprocal slopes of [tex][tex]$-\frac{1}{3}$[/tex][/tex] and 3 and are, therefore, perpendicular.
D. They have the same slope of [tex][tex]$\frac{1}{3}$[/tex][/tex] and are, therefore, parallel.

Sagot :

GinnyAnswer
To determine the relationship between segments [tex]\(XY\)[/tex] and [tex]\(WZ\)[/tex], we need to analyze the slopes of the lines that contain these segments based on their given equations.

1. Convert the equations of the lines to slope-intercept form [tex]\( y = mx + b \)[/tex], where [tex]\( m \)[/tex] is the slope.

- For the line [tex]\( X-3y=-12 \)[/tex]:
[tex]\[ X - 3y = -12 \\ -3y = -X - 12 \\ y = \frac{1}{3}X + 4 \][/tex]
The slope [tex]\( m \)[/tex] of this line is [tex]\( \frac{1}{3} \)[/tex].

- For the line [tex]\( X-3y=-6 \)[/tex]:
[tex]\[ X - 3y = -6 \\ -3y = -X - 6 \\ y = \frac{1}{3}X + 2 \][/tex]
The slope [tex]\( m \)[/tex] of this line is also [tex]\( \frac{1}{3} \)[/tex].

2. Compare the slopes.

- The slope of the line containing segment [tex]\(XY\)[/tex] is [tex]\( \frac{1}{3} \)[/tex].
- The slope of the line containing segment [tex]\(WZ\)[/tex] is [tex]\( \frac{1}{3} \)[/tex].

3. Determine the relationship.

Since both lines have the same slope of [tex]\( \frac{1}{3} \)[/tex], they are parallel.

Therefore, the correct statement is:

They have the same slope of [tex]\(\frac{1}{3}\)[/tex] and are, therefore, parallel.
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