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Are these lines perpendicular, parallel, or neither based on their slopes?
[tex]$
\begin{array}{l}
6x - 2y = -2 \\
y = 3x + 12
\end{array}
$[/tex]

The [tex]$\square$[/tex] of their slopes is [tex]$\square$[/tex], so the lines are [tex]$\square$[/tex].



Answer :

Let's analyze the given equations to determine the nature of the lines:

1. First Equation: [tex]\(6x - 2y = -2\)[/tex]
- To rewrite this in the slope-intercept form [tex]\(y = mx + b\)[/tex], we need to solve for [tex]\(y\)[/tex]:
[tex]\[ 6x - 2y = -2 \][/tex]
[tex]\[ -2y = -6x - 2 \][/tex]
[tex]\[ y = 3x + 1 \][/tex]

Therefore, the slope ([tex]\(m\)[/tex]) of the first equation is [tex]\(3\)[/tex].

2. Second Equation: [tex]\(y = 3x + 12\)[/tex]
- This is already in the slope-intercept form [tex]\(y = mx + b\)[/tex], where [tex]\(m = 3\)[/tex].

Now, we have both lines with their slopes:
- First line: [tex]\(y = 3x + 1\)[/tex] with slope [tex]\(3\)[/tex]
- Second line: [tex]\(y = 3x + 12\)[/tex] with slope [tex]\(3\)[/tex]

3. Relationship Between the Lines:
- We compare the slopes of the two lines. If two lines have the same slope and different y-intercepts, they are parallel.
- Here, both slopes are [tex]\(3\)[/tex].

Since the slopes of both lines are equal ([tex]\(3\)[/tex]), the lines are parallel.

Therefore, the correct answers for the statement are:

- The slope of their slopes is 3 (for both lines), so the lines are parallel.