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Are these lines perpendicular, parallel, or neither based on their slopes?
[tex]\[6x - 2y = -2\][/tex]
[tex]\[y = 3x + 12\][/tex]

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



Answer :

GinnyAnswer
To determine whether the lines represented by the equations [tex]\( 6x - 2y = -2 \)[/tex] and [tex]\( y = 3x + 12 \)[/tex] are perpendicular, parallel, or neither, follow these steps:

1. Convert the first equation to slope-intercept form [tex]\( y = mx + b \)[/tex].
- Start with the equation: [tex]\( 6x - 2y = -2 \)[/tex].
- Solve for [tex]\( y \)[/tex]:
[tex]\[ -2y = -6x - 2 \][/tex]
Divide every term by -2:
[tex]\[ y = 3x + 1 \][/tex]
- The slope ([tex]\( m \)[/tex]) of the first line is 3.

2. The slope of the second line [tex]\( y = 3x + 12 \)[/tex] is already given as the coefficient of [tex]\( x \)[/tex].
- Thus, the slope ([tex]\( m \)[/tex]) of the second line is also 3.

3. Compare the slopes to determine the relationship between the lines:
- If the slopes are equal, the lines are parallel.
- If the product of the slopes is -1, the lines are perpendicular.
- If neither condition is satisfied, the lines are neither parallel nor perpendicular.

In this case, both lines have a slope of 3. Since the slopes are the same, the lines are parallel.

Therefore:

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

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