Given:
The equation of line c is,
[tex]y=-6x-3[/tex]Line d is perpendicular to line c, that means their slopes will be opposite reciprocals.
Slope of line c is,
[tex]\begin{gathered} \text{For y=mx+c} \\ m=\text{ slope} \\ y=-6x-3 \\ \text{slope =m}_1=-6 \end{gathered}[/tex]It gives,
[tex]\begin{gathered} m_1\times m_2=-1 \\ -6\times m_2=-1_{} \\ m_2=-\frac{1}{-6}=\frac{1}{6} \end{gathered}[/tex]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,
[tex]\begin{gathered} y-y_1=m_2(x-x_1) \\ (x_1,y_1)=(3,-3) \\ y-(-3)=\frac{1}{6}(x-3) \\ y+3=\frac{1}{6}x-\frac{3}{6} \\ y=\frac{1}{6}x-\frac{1}{2}-3 \\ y=\frac{1}{6}x-\frac{7}{2} \end{gathered}[/tex]Answer: the equation of line d is,
[tex]y=\frac{1}{6}x-\frac{7}{2}[/tex]
