Find the equation of the line parallel to [tex] y = 5x + 2 [/tex] that passes through the point [tex] (-6, -1) [/tex].

A. [tex] y = -5x - 11 [/tex]
B. [tex] y = 5x + 29 [/tex]
C. [tex] y = \frac{1}{5}x + \frac{1}{6} [/tex]
D. [tex] y = 5x - 29 [/tex]



Answer :

To find the equation of a line parallel to [tex]\( y = 5x + 2 \)[/tex] that passes through the point [tex]\((-6, -1)\)[/tex], we need to follow these steps:

1. Identify the slope of the original line:
The given line is in the slope-intercept form [tex]\( y = mx + c \)[/tex], where [tex]\( m \)[/tex] is the slope. For the equation [tex]\( y = 5x + 2 \)[/tex], the slope [tex]\( m \)[/tex] is 5.

2. Determine the slope of the parallel line:
Two lines are parallel if they have the same slope. Therefore, the slope of the required parallel line is also 5.

3. Use the point-slope form to find the equation:
The point-slope form of a line's equation is given by [tex]\( y - y_1 = m(x - x_1) \)[/tex], where [tex]\( (x_1, y_1) \)[/tex] is a point on the line and [tex]\( m \)[/tex] is the slope.
Given the point [tex]\((-6, -1)\)[/tex] and the slope [tex]\( 5 \)[/tex], we can substitute these values into the point-slope form:
[tex]\[ y - (-1) = 5(x - (-6)) \][/tex]
Simplifying, we get:
[tex]\[ y + 1 = 5(x + 6) \][/tex]

4. Expand and simplify to get the slope-intercept form:
Next, simplify the equation:
[tex]\[ y + 1 = 5x + 30 \][/tex]
Subtract 1 from both sides to solve for [tex]\( y \)[/tex]:
[tex]\[ y = 5x + 29 \][/tex]

Thus, the equation of the line parallel to [tex]\( y = 5x + 2 \)[/tex] that passes through the point [tex]\((-6, -1)\)[/tex] is:
[tex]\[ y = 5x + 29 \][/tex]

Therefore, the correct answer is:
[tex]\[ \boxed{y = 5x + 29} \][/tex]

From the given choices, the answer is:
[tex]\[ \boxed{B} \][/tex]