We want to find the equation of a line with slope -2/3 and goes through the point (-6,-5).
We can write the general equation of a line as:
[tex]\begin{gathered} y=mx+b \\ \text{Where m is the slope and b is the y-intercept value} \end{gathered}[/tex]In this case, m=-2/3 and the point (-6,-5) must satisfy the equation, so:
[tex]\begin{gathered} y=-\frac{2}{3}x+b \\ We\text{ find the value of b with the point (x,y)=(-6,-5):} \\ -5=-\frac{2}{3}\cdot(-6)+b \\ -5=\frac{12}{3}+b \\ -5=4+b \\ b=-5-4=-9 \end{gathered}[/tex]Now, we replace the value of b and write the equation in the standard form:
[tex]\begin{gathered} y=-\frac{2}{3}x-9 \\ We\text{ can multiply al the equation by 3:} \\ 3\cdot y=3\cdot(-\frac{2}{3}x-9) \\ 3y=-\frac{2}{3}\cdot3\cdot x-9\cdot3 \\ 3y=-2x-27 \\ 2x+3y=-27 \end{gathered}[/tex]So, the option A is the correct answer.
