The line is perpendicular to the line:
![x-7y=-3](https://img.qammunity.org/2023/formulas/mathematics/college/vt3cipsh99152gfjyftgourwdpjamrm8id.png)
We can rewrite the equation in the slope-intercept form to be:
![\begin{gathered} 7y=x+3 \\ y=(1)/(7)x+(3)/(7) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/qcphm8z0l90eylg6uolma8r1y77gkjf3fx.png)
This equation compared to the slope-intercept form will give the slope as follows:
![\begin{gathered} y=mx+b,m=slope \\ \therefore \\ m=(1)/(7) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/8dzpwlszv96d0l5z111diafujd5ghdv190.png)
Recall that perpendicular lines have slopes that are negative reciprocals. Thus:
![m_1=-(1)/(m_2)](https://img.qammunity.org/2023/formulas/mathematics/college/4qv83ntgiq6ef6mdrf1mcjqzd45udgvi5w.png)
Hence, the slope of the required line will be:
![m=-7](https://img.qammunity.org/2023/formulas/mathematics/college/udy3rxbykk64n05jqszs0q2zsk9qb9xb11.png)
Given that we have the point the line passes through given, we can put the equation in the point-slope form:
![y-y_1=m(x-x_1)](https://img.qammunity.org/2023/formulas/mathematics/high-school/csobd57zth7rh9k4hz9amldzpq2owf0z4j.png)
At the point (0, 8), we have the equation to be:
![\begin{gathered} y-8=-7(x-0) \\ y-8=-7x \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/q4ttjjnrh5imwt3bi8anmvpveoho35xppt.png)
In the slope-intercept form, the equation of the line will be:
![y=-7x+8](https://img.qammunity.org/2023/formulas/mathematics/college/r2z4edxl0chkkf5r4whov80l1ju4f5vevk.png)