Write the equation of a line that is parallel to [tex] y = 0.6x + 3 [/tex] and passes through the point [tex] (-3, -5) [/tex].

[tex] \boxed{} [/tex]



Answer :

To find the equation of a line that is parallel to [tex]\( y = 0.6x + 3 \)[/tex] and passes through the point [tex]\((-3, -5)\)[/tex], follow these steps:

1. Identify the slope of the given line:
The given line has the equation [tex]\( y = 0.6x + 3 \)[/tex]. The coefficient of [tex]\( x \)[/tex] in this equation, which is [tex]\( 0.6 \)[/tex], is the slope of the line. Therefore, the slope ([tex]\( m \)[/tex]) of the given line is [tex]\( 0.6 \)[/tex].

2. Determine the slope of the parallel line:
Lines that are parallel have the same slope. Thus, the slope of our new line is also [tex]\( 0.6 \)[/tex].

3. Use the point-slope form of the equation:
The point-slope form of a line's equation is [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.
We are given the point [tex]\((-3, -5)\)[/tex] and the slope [tex]\( 0.6 \)[/tex]. Substituting these values into the point-slope form:

[tex]\[ y - (-5) = 0.6(x - (-3)) \][/tex]

4. Simplify the equation:
Simplify the point-slope equation to convert it to the slope-intercept form [tex]\( y = mx + b \)[/tex]:

[tex]\[ y + 5 = 0.6(x + 3) \][/tex]
Distribute the slope [tex]\( 0.6 \)[/tex]:

[tex]\[ y + 5 = 0.6x + 1.8 \][/tex]

Isolate [tex]\( y \)[/tex] by subtracting [tex]\( 5 \)[/tex] from both sides:

[tex]\[ y = 0.6x + 1.8 - 5 \][/tex]

Simplify the right side:

[tex]\[ y = 0.6x - 3.2 \][/tex]

Therefore, the equation of the line parallel to [tex]\( y = 0.6x + 3 \)[/tex] and passing through the point [tex]\((-3, -5)\)[/tex] is:

[tex]\[ y = 0.6x - 3.2 \][/tex]