## Algebra: A Combined Approach (4th Edition)

$5x-8y=20$
The equation of a line is: $y-y_{1}=m(x-x_{1})$ But we know that: $m=\dfrac{y_{2}-y_{1}}{x_{2}-x_{1}}$ Thus, substituting we have: $y-y_{1}=\dfrac{y_{2}-y_{1}}{x_{2}-x_{1}}(x-x_{1})$ Given points $(4,0)$ and $(-4,-5)$ we have that: $y_{1}=-5 ,\ x_{1}=-4,\ y_{2}=0,\ x_{2}=4$ Substituting, we have: $y-(-5)=\dfrac{0-(-5)}{4-(-4)}(x-(-4)) \longrightarrow y+5=\dfrac{0+5}{4+4}(x+4) \longrightarrow \\ y+5=\dfrac{5}{8}(x+4) \longrightarrow y=\dfrac{5}{8}x+4\times \dfrac{5}{8} - 5 \longrightarrow y=\dfrac{5}{8}x+\dfrac{5}{2} - 5 \longrightarrow \\ y=\dfrac{5}{8}x-\dfrac{5}{2} \longrightarrow \dfrac{5}{2}=\dfrac{5}{8}x-y \longrightarrow 8\times \dfrac{5}{2}=8\times \dfrac{5}{8}x-8\times y \longrightarrow 20=5x-8y \longrightarrow 5x-8y=20$