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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 :

Let's analyze the problem step-by-step to determine whether the given lines are perpendicular, parallel, or neither based on their slopes.

First, we need to identify the slopes of each line.

1. Equation 1: [tex]\( 6x - 2y = -2 \)[/tex]
- To find the slope of this line, we need to rearrange the equation into the slope-intercept form: [tex]\( y = mx + b \)[/tex], where [tex]\( m \)[/tex] is the slope.
- Starting with [tex]\( 6x - 2y = -2 \)[/tex]:
[tex]\[ 6x - 2y = -2 \][/tex]
- Isolate [tex]\( y \)[/tex] by moving [tex]\( 6x \)[/tex] to the right side:
[tex]\[ -2y = -6x - 2 \][/tex]
- Divide every term by -2 to solve for [tex]\( y \)[/tex]:
[tex]\[ y = 3x + 1 \][/tex]
- Therefore, the slope ([tex]\( m \)[/tex]) of the first line is [tex]\( 3 \)[/tex].

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

Now that we have the slopes of both lines:

- Slope of the first line ([tex]\( m_1 \)[/tex]): [tex]\( 3 \)[/tex]
- Slope of the second line ([tex]\( m_2 \)[/tex]): [tex]\( 3 \)[/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 [tex]\( -1 \)[/tex] ([tex]\( m_1 \cdot m_2 = -1 \)[/tex]), the lines are perpendicular.
- If neither condition is met, the lines are neither parallel nor perpendicular.

Given that both slopes are equal ([tex]\( 3 = 3 \)[/tex]), the lines are parallel.

As a result, the correct answer is:

The product of their slopes is 9 (since [tex]\( 3 \times 3 = 9 \)[/tex]),
so the lines are parallel.
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