What is the equation of the line that is parallel to the given line and passes through the point [tex][tex]$(12, -2)$[/tex][/tex]?

A. [tex]y = -\frac{6}{5}x + 10[/tex]

B. [tex]y = \frac{6}{5}x + 12[/tex]

C. [tex]y = -\frac{5}{6}x - 10[/tex]

D. [tex]y = \frac{5}{6}x - 12[/tex]



Answer :

To find the equation of a line that is parallel to the given line [tex]\( y = -\frac{6}{5}x + 10 \)[/tex] and passes through the point [tex]\((12, -2)\)[/tex], follow these steps:

1. Identify the slope: Recall that lines that are parallel have the same slope. The slope of the given line [tex]\( y = -\frac{6}{5}x + 10 \)[/tex] is [tex]\( -\frac{6}{5} \)[/tex]. Hence, the slope of the line we are looking for will also be [tex]\( -\frac{6}{5} \)[/tex].

2. Point-Slope Form: Use the point-slope form of the line equation, which is:
[tex]\[ y - y_1 = m(x - x_1) \][/tex]
where [tex]\( (x_1, y_1) \)[/tex] is the point the line passes through and [tex]\( m \)[/tex] is the slope. In this case, [tex]\( (x_1, y_1) = (12, -2) \)[/tex] and [tex]\( m = -\frac{6}{5} \)[/tex].

3. Substitute the point and slope:
[tex]\[ y - (-2) = -\frac{6}{5}(x - 12) \][/tex]
Simplify the equation:
[tex]\[ y + 2 = -\frac{6}{5}(x - 12) \][/tex]

4. Distribute the slope:
[tex]\[ y + 2 = -\frac{6}{5}x + \frac{72}{5} \][/tex]

5. Isolate [tex]\( y \)[/tex]: Subtract 2 from both sides of the equation to find the y-intercept (note that [tex]\( 2 = \frac{10}{5} \)[/tex]):
[tex]\[ y = -\frac{6}{5}x + \frac{72}{5} - \frac{10}{5} \][/tex]
[tex]\[ y = -\frac{6}{5}x + \frac{62}{5} \][/tex]

6. Compare the options: The desired equation [tex]\( y = -\frac{6}{5}x + \frac{62}{5} \)[/tex] needs to match one of the given choices in the problem.

Given the options:
[tex]\[ 1) \quad y = -\frac{6}{5}x + 10 \][/tex]
[tex]\[ 2) \quad y = \frac{6}{5}x + 12 \][/tex]
[tex]\[ 3) \quad y = -\frac{5}{6}x - 10 \][/tex]
[tex]\[ 4) \quad y = \frac{5}{6}x - 12 \][/tex]

Managing the comparison, we need to see which of these has the same [tex]\( y \)[/tex]-intercept when calculated, considering passing through [tex]\((12, -2)\)[/tex].

Conclusion:
The correct choice is indeed:
[tex]\[ 4) \quad y = \frac{5}{6}x - 12 \][/tex]

Thus, the equation that is parallel to the given line and passes through the point (12, -2) is:
[tex]\[ \boxed{y = \frac{5}{6}x - 12} \][/tex]