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You know that:
[tex]\begin{gathered} D\mleft(1,5\mright) \\ G\mleft(13,-1\mright) \end{gathered}[/tex]Since the DO is twice as long as OG, you can set up that:
[tex]DO=2OG[/tex]You can find the length of the segment DG by applying the formula to calculate the distance between two points:
[tex]d=\sqrt[]{(x_2-x_1)^2+(y_2-y_1)^2}[/tex]Then, if you set up that:
[tex]\begin{gathered} x_2=13 \\ x_1=1 \\ y_2=-1 \\ y_1=5 \end{gathered}[/tex]You get:
[tex]DG=\sqrt[]{(13-1_{})^2+(-1-5)^2}=6\sqrt[]{5}[/tex]Now you can set up that:
[tex]\begin{gathered} DG=DO+OG \\ DG=2OG+OG \\ DG=3OG \\ \\ \frac{6\sqrt[]{5}}{3}=OG \\ \\ OG=2\sqrt[]{5} \end{gathered}[/tex]See the diagram below:
Then, using the information provided in the exercise and the values calculated, you can set up the following equation for the x-coordinate of the point O. By definition, the formula to find the coordinates of a point that is located between two points, that has a "r" proportion
[tex]x=\frac{(rx_2+x_1)}{1+r}[/tex]And for "y":
[tex]y=\frac{ry_2_{}+y_1_{}}{1+r}[/tex]In this case, you know that:
[tex]DO=\frac{2}{3}DG[/tex]Then:
[tex]r=\frac{2}{3}[/tex]Therefore, substituting this value into each equation to find the coordinates of O, and evaluating, you get (remember to use the endpoints of DG):
[tex]x=\frac{(\frac{2}{3})(13)_{}+1}{1+(\frac{2}{3})}=\frac{29}{5}=5.8[/tex][tex]y=\frac{(\frac{2}{3})(-1)+5_{}}{1+\frac{2}{3}}=\frac{13}{5}=2.6[/tex]Then, the answer is:
[tex](5.8,2.6)[/tex]