There are two ways to divide segment AB into two subsegments with a 2:1 ratio
[tex]\begin{gathered} AC=2CB \\ \text{and} \\ 2AC=CB \end{gathered}[/tex]
In general, the distance between two points on the plane is given by the formula below
[tex]\begin{gathered} X=(x_1,y_1),Y=(x_2,y_2) \\ \Rightarrow d(X,Y)=\sqrt[]{(x_1-x_2)^2+(y_1-y_2)^2} \end{gathered}[/tex]
Furthermore, point C has to be on the same line as A and B, whose equation is
[tex]\begin{gathered} y-3=\frac{-3-3}{6-(-3)}(x-(-3)) \\ \Rightarrow y-3=-\frac{2}{3}(x+3) \\ \Rightarrow y=-\frac{2}{3}x+1 \end{gathered}[/tex]
Additionally, the distance between A and B is
[tex]d(A,B)=\sqrt[]{(-9)^2+(6)^2}=\sqrt[]{117}[/tex]
1) AC=2CB
[tex]\begin{gathered} d(C,B)=\frac{\sqrt[]{117}}{3} \\ \Rightarrow\sqrt[]{(x-6)^2+(y+3)^2}=\frac{\sqrt[]{117}}{3} \\ \Rightarrow(x-6)^2+(y+3)^2=\frac{117}{9} \end{gathered}[/tex]
On the other hand, using the equation of the line,
[tex]\begin{gathered} y=-\frac{2}{3}x+1 \\ \Rightarrow(x-6)^2+(-\frac{2}{3}x+4)^2=\frac{117}{3} \\ \Rightarrow\frac{13}{9}(x-6)^2=\frac{117}{3} \\ \Rightarrow(x-6)^2=27 \\ \Rightarrow x-6=\pm\sqrt[]{27} \\ \Rightarrow x=6\pm\sqrt[]{27} \end{gathered}[/tex]
However, x=6+sqrt(27)=11.19...-> out of the segment; therefore, the only valid value of x is
[tex]x=6-\sqrt[]{27}[/tex]
Finding y,
