1. The problem asks to find the standard and general form of the line passing through the points (-8,-6) and (-16,2). 2. First, find the slope $m$ using the formula: $$m=\frac{y_2-y_1}{x_2-x_1}$$ where $(x_1,y_1)=(-8,-6)$ and $(x_2,y_2)=(-16,2)$. 3. Calculate the slope: $$m=\frac{2-(-6)}{-16-(-8)}=\frac{2+6}{-16+8}=\frac{8}{-8}=-1$$ 4. Use the point-slope form of a line: $$y-y_1=m(x-x_1)$$ Substitute $m=-1$ and point $(-8,-6)$: $$y-(-6)=-1(x-(-8))$$ $$y+6=-1(x+8)$$ 5. Simplify the equation: $$y+6=-x-8$$ 6. Rearrange to get the standard form $Ax+By=C$: $$x+y=-14$$ 7. The general form is the same as the standard form but with all terms on one side: $$x+y+14=0$$ Final answers: - Standard form: $$x+y=-14$$ - General form: $$x+y+14=0$$