Given:
The equation of line c is,
![y=-6x-3](https://img.qammunity.org/2023/formulas/mathematics/college/drgoofqxax3apb3htrz3ax5mhq1cu62aei.png)
Line d is perpendicular to line c, that means their slopes will be opposite reciprocals.
Slope of line c is,
![\begin{gathered} \text{For y=mx+c} \\ m=\text{ slope} \\ y=-6x-3 \\ \text{slope =m}_1=-6 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/nmpw0dawsyo852ygk6xixmxwzrvsiwnsxy.png)
It gives,
![\begin{gathered} m_1* m_2=-1 \\ -6* m_2=-1_{} \\ m_2=-(1)/(-6)=(1)/(6) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/oejcbyc4qk75mz9hy6u5q2c2ycgxmhuc3h.png)
So, slope of the line d is 1/6.
Also line d includes point ( 3, -3 )
The point slope form of equation of line is,
![\begin{gathered} y-y_1=m_2(x-x_1) \\ (x_1,y_1)=(3,-3) \\ y-(-3)=(1)/(6)(x-3) \\ y+3=(1)/(6)x-(3)/(6) \\ y=(1)/(6)x-(1)/(2)-3 \\ y=(1)/(6)x-(7)/(2) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/s1kvjdzsiv8n8hhiwcpejjoonteeyxfd9p.png)
Answer: the equation of line d is,
![y=(1)/(6)x-(7)/(2)](https://img.qammunity.org/2023/formulas/mathematics/college/h9ap3jcufg7tljsx95ahzxj8duicv0hzwb.png)