The point-slope form of a line:
[tex]y-y_1=m(x-x_1)\\\\m=\dfrac{y_2-y_1}{x_2-x_1}[/tex]
We have the points (-9, -5) and (2, -4). Substitute:
[tex]m=\dfrac{-4-(-5)}{2-(-9)}=\dfrac{1}{11}\\\\y-(-5)=\dfrac{1}{11}(x-(-9))\\\\\boxed{y+5=\dfrac{1}{11}(x+9)}\leftarrow\boxed{\text{point-slope form}}[/tex]
The standard form of a line:
[tex]Ax+By=C[/tex]
[tex]y+5=\dfrac{1}{11}(x+9)[/tex] multiply both sides by 11
[tex]11y+55=1(x+9)[/tex]
[tex]11y+55=x+9[/tex] subtract 9 from both sides
[tex]11y+46=x[/tex] subtract 11y from both sides
[tex]46=x-11y[/tex]
[tex]\boxed{x-11y=46}\leftarrow\boxed{\text{standard form}}[/tex]
