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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].

Sagot :

GinnyAnsweridn
To determine whether the lines are perpendicular, parallel, or neither, we need to analyze their slopes. Let's first convert each equation to the slope-intercept form, [tex]\( y = mx + b \)[/tex], where [tex]\( m \)[/tex] represents the slope.

1. For the first equation [tex]\( 6x - 2y = -2 \)[/tex]:

- Rewrite it to solve for [tex]\( y \)[/tex]:
[tex]\[ 6x - 2y = -2 \][/tex]
[tex]\[ -2y = -6x - 2 \][/tex]
[tex]\[ y = 3x + 1 \][/tex]
- The slope [tex]\( m_1 \)[/tex] of the first line is [tex]\( 3 \)[/tex].

2. The second equation, [tex]\( y = 3x + 12 \)[/tex], is already in slope-intercept form.

- The slope [tex]\( m_2 \)[/tex] of the second line is also [tex]\( 3 \)[/tex].

Now, we compare the slopes [tex]\( m_1 \)[/tex] and [tex]\( m_2 \)[/tex]:

- If the slopes are equal, the lines are parallel.
- If the product of the slopes is [tex]\(-1\)[/tex], the lines are perpendicular.
- If neither condition is met, the lines are neither parallel nor perpendicular.

Here, the slopes are:

- [tex]\( m_1 = 3 \)[/tex]
- [tex]\( m_2 = 3 \)[/tex]

Since [tex]\( m_1 = m_2 \)[/tex]:

- The lines are parallel.

Therefore, we can complete the statement as follows:

The comparison of their slopes is equal, so the lines are parallel.
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