The equation of line passing through (-8, -2) and (-4, 6) is y = 2x + 14Solution:Given that Line is passing through point (− 8 ,− 2) and ( -4 , 6 )Equation of line passing through point [tex]\left(x_{1}, y_{1}\right) \text { and }\left(x_{2}, y_{2}\right)[/tex] is given by:[tex]y-y_{1}=\frac{\left(y_{2}-y_{1}\right)}{\left(x_{2}-x_{1}\right)}\left(x-x_{1}\right)[/tex] ----- eqn 1[tex]\text { In our case } x_{1}=-8, y_{1}=-2, x_{2}=-4, y_{2}=6[/tex]Substituting given value in (1) we get[tex]\begin{array}{l}{y-(-2)=\frac{(6-(-2))}{(-4-(-8))}(x-(-8))} \\\\ {=>y+2=\frac{8}{4}(x+8)} \\\\ {=>y+2=2(x+8)} \\\\ {=>y+2=2 x+16} \\\\ {=>-2 x+y=14}\end{array}[/tex]Thus the required equation of line is y = 2x + 14