Two parallel lines have the same slope. Given the equation of one of the lines, you can determine the slope of the other line:
![y=-5x+1](https://img.qammunity.org/2023/formulas/mathematics/high-school/1bapp6rwwmrds98uw4z7gfgbbhsu2cah5x.png)
The slope of the line is the coefficient of the x-term, in this case, that coefficient is -5. Then the slope of both parallel lines is m= -5.
The line you have to find must cross through the point (9,-42). Using the point-slope form you can determine the equation of the parallel line:
![y-y_1=m(x-x_1)](https://img.qammunity.org/2023/formulas/mathematics/high-school/csobd57zth7rh9k4hz9amldzpq2owf0z4j.png)
Where
m is the slope of the line
(x₁,y₁) are the coordinates of one point of the line:
![\begin{gathered} y-(-42)=-5(x-9) \\ y+42=-5(x-9) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/high-school/7iclbi69cq8i8r85wifqn1pz6wwk9ou16i.png)
To write the equation in slope-intercept form, the first step is to distribute the multiplication on the parentheses term:
![\begin{gathered} y+42=(-5)\cdot x+(-5)\cdot(-9) \\ y+42=-5x+45 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/high-school/rlcfd7jt5vgrqm83mul8ziosls0xihcsio.png)
Subtract 42 to both sides of the expression to pass the term to the right side of the equal sign:
![\begin{gathered} y+42-42=-5x+45-42 \\ y=-5x+3 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/high-school/hdpmpqmxiq342b2baoqa8yxqu96h46efid.png)
The equation of the line is y= -5x + 3