Questions: Consider the line 6x-2y=-1. Find the equation of the line that is perpendicular to this line and passes through the point (-8,1). Find the equation of the line that is parallel to this line and passes through the point (-8,1). Equation of perpendicular line: Equation of parallel line:

Solution Steps

The given line is represented by the equation \( 6x - 2y = -1 \). To find the slope of this line, we can rearrange it into slope-intercept form \( y = mx + c \):

\[ 6x - 2y = -1 \implies 2y = 6x + 1 \implies y = 3x + \frac{1}{2} \]

The slope \( m \) of the given line is \( 3 \). The slope of the line that is perpendicular to this line is the negative reciprocal of \( 3 \), which is \( -\frac{1}{3} \).

Next, we need to find the equation of the perpendicular line that passes through the point \( (-8, 1) \). Using the point-slope form of the line equation:

\[ y - y_1 = m(x - x_1) \]

Substituting \( m = -\frac{1}{3} \), \( x_1 = -8 \), and \( y_1 = 1 \):

\[ y - 1 = -\frac{1}{3}(x + 8) \]

Solving for \( y \):

\[ y = -\frac{1}{3}x - \frac{8}{3} + 1 = -\frac{1}{3}x - \frac{5}{3} \]

Thus, the equation of the perpendicular line is:

\[ \boxed{y = -\frac{1}{3}x - \frac{5}{3}} \]

The slope of the parallel line will be the same as the original line, which is \( 3 \). We need to find the equation of the parallel line that also passes through the point \( (-8, 1) \).

Using the point-slope form again:

\[ y - 1 = 3(x + 8) \]

Solving for \( y \):

\[ y - 1 = 3x + 24 \implies y = 3x + 25 \]

Thus, the equation of the parallel line is:

\[ \boxed{y = 3x + 25} \]

Final Answer