The equation of the line passing through point $ \left(4,~-5\right) $, and point $ \left(6,~-13\right) $ is:
$$ 4x+y-11=0 $$To find equation of the line passing through points $ A(x_A,y_A) $ and $ B(x_B,y_B) $, we use formula:
$$ y - y_A~=~\frac{y_B - y_A}{x_B - x_A}(x-x_A) $$In this example we have:
$$ \begin{aligned} & \left(4,~-5\right) \implies x_A = 4 ~~\text{and}~~ y_A = -5 \\[1 em] & \left(6,~-13\right) \implies x_B = 6 ~~\text{and}~~ y_B = -13 \end{aligned} $$After substituting into the formula, we obtain:
$$ \begin{aligned} y - y_A~&=~\frac{y_B - y_A}{x_B - x_A}(x - x_A) \\[1 em] y - \left(-5\right)~&=~\frac{ -13 - \left(-5\right) }{ 6 - 4 } \left( x - 4 \right) \\[1 em]y + 5 ~&=~ -4 \left( x - 4 \right) \\[1 em]y + 5 ~&=~ -4x + 16 \\[1 em]y ~&=~ -4x + 11 \end{aligned} $$