Solution:
To find the equation of a line, the formula is
[tex]\frac{y-y_1}{x-x_1}=\frac{y_2-y_1}{x_2-x_1}[/tex]Where
[tex]\begin{gathered} (x_1,y_1)=(3,-2) \\ (x_2,y_2)=(-4,8) \end{gathered}[/tex]Substitute the values of the variable
[tex]\begin{gathered} \frac{y-(-2)}{x-3}=\frac{-8-(-2)}{-4-3} \\ \frac{y+2}{x-3}=\frac{-8+2}{-7} \\ \frac{y+2}{x-3}=\frac{-6}{-7} \end{gathered}[/tex][tex]\begin{gathered} Crossmultiply \\ -7(y+2)=-6(x-3) \\ -7y-14=-6x+18 \\ 6x-7y-18-14=0 \\ 6x-7y-32=0 \end{gathered}[/tex]Hence, the equation is
[tex]6x-7y-32=0[/tex]