Solution:
Given that;
Line 1 passes through the points A(-15,-8) and B(-3,0)
To find the slope, m, of the line, the formula is
[tex]\begin{gathered} m=\frac{y_2-y_1}{x_2-x_1} \\ (x_1,y_1)=(-15,-8) \\ (x_2,y_2)=(-3,0) \end{gathered}[/tex]Substituting the coordinates to find the slope of line 1
[tex]\begin{gathered} m=\frac{0-(-8)}{-3-(-15)}=\frac{8}{-3+15}=\frac{8}{12}=\frac{2}{3} \\ m=\frac{2}{3} \end{gathered}[/tex]Since, line 3 is parallel to line 1, then, they will have the same slope,
Thus, the slope pf line 3 is 2/3
Line 2 has equation 3x + 2y - 35 = 0.
Where line 3 intersects line 2 at x = 5, substitute for x into the equation of line 2 to find the value of y
[tex]\begin{gathered} 3x+2y-35=0 \\ 3(5)+2y-35=0 \\ 15+2y-35=0 \\ -20=-2y \\ y=\frac{-10}{-2}=5 \\ y=5 \end{gathered}[/tex]Where
The slope and the coordinates on line 3 are
[tex]\begin{gathered} m=\frac{2}{3} \\ (x_1,y_1)=(5,10) \end{gathered}[/tex]Applying the point-slope formula to find the equation of line 2
[tex]\begin{gathered} y-y_1=m(x-x_1) \\ y-10=\frac{2}{3}(x-5) \\ 3(y-10)=2(x-5) \\ 3y-30=2x-10 \\ 2x-10-(3y-30)=0 \\ 2x-10-3y+30=0 \\ 2x-3y+20=0 \end{gathered}[/tex]Hence, the equation of line 3 is
[tex]2x-3y+20=0[/tex]