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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 :

To determine the relationship between the two lines, we need to find their slopes and compare them. Here, we have the following equations:

1. [tex]\( 6x - 2y = -2 \)[/tex]
2. [tex]\( y = 3x + 12 \)[/tex]

First, let's put the first equation into the slope-intercept form [tex]\( y = mx + b \)[/tex], where [tex]\( m \)[/tex] represents the slope. Starting with the given equation:

[tex]\[ 6x - 2y = -2 \][/tex]

Solve for [tex]\( y \)[/tex]:

[tex]\[ -2y = -6x - 2 \][/tex]

[tex]\[ y = 3x + 1 \][/tex]

From this, we can see that the slope ([tex]\( m_1 \)[/tex]) of the first line is 3.

The second equation is already in slope-intercept form [tex]\( y = mx + b \)[/tex]. From [tex]\( y = 3x + 12 \)[/tex], we can directly read off the slope ([tex]\( m_2 \)[/tex]), which is also 3.

Now, we have the slopes [tex]\( m_1 = 3 \)[/tex] and [tex]\( m_2 = 3 \)[/tex].

To determine the relationship between the lines:
- If the slopes are equal ([tex]\( m_1 = m_2 \)[/tex]), the lines are parallel.
- If the product of the slopes is -1 ([tex]\( m_1 \cdot m_2 = -1 \)[/tex]), the lines are perpendicular.
- If neither condition is met, the lines are neither parallel nor perpendicular.

In this case, [tex]\( m_1 = 3 \)[/tex] and [tex]\( m_2 = 3 \)[/tex], so the slopes are equal. Thus, the lines are parallel.

Therefore, we can fill in the blanks as follows:

The [tex]\(\boxed{\text{slope}}\)[/tex] of their slopes is [tex]\(\boxed{\text{equal}}\)[/tex], so the lines are [tex]\(\boxed{\text{parallel}}\)[/tex].