Answer :
A line segment has the following endpoints
[tex]P(1,-1)\: and\: Q(-9,-6)[/tex]We are asked to find the coordinates of point R, such that the point R will be 3/5 of the distance PQ.
Let us first find the distance from P to Q.
The distance between two points P and Q is given by
[tex]PQ=\sqrt[]{\mleft({x_2-x_1}\mright)^2+\mleft({y_2-y_1}\mright)^2}[/tex]Let us substitute the given points into the above distance formula
[tex]\begin{gathered} PQ=\sqrt[]{({-9_{}-1_{}})^2+({-6_{}-(-1)_{}})^2} \\ PQ=\sqrt[]{({-9_{}-1_{}})^2+({-6_{}+1_{}})^2} \\ PQ=\sqrt[]{({-10})^2+({-5})^2} \\ PQ=\sqrt[]{100^{}+25^{}} \\ PQ=\sqrt[]{125} \\ PQ=11.18 \end{gathered}[/tex]Now let us find the coordinates of point R such that R is equal to 3/5 of PQ.
So the ratio is m:n = 3:5
[tex]R_x=\frac{nx_1+mx_2}{n+m},R_y=\frac{ny_1+my_2}{n+m}[/tex]Let us substitute the given values into the above formula
[tex]\begin{gathered} R_x=\frac{5\cdot1+3\cdot(-9)}{3+5},R_y=\frac{5\cdot(-1)+3\cdot(-6)}{3+5} \\ R_x=\frac{5-27}{8},R_y=\frac{-5-18}{8} \\ R_x=\frac{22}{8},R_y=\frac{-23}{8} \\ R_x=\frac{11}{4},R_y=\frac{-23}{8} \end{gathered}[/tex]Therefore, the coordinates of the point R is
[tex]R(\frac{11}{4},-\frac{23}{8})[/tex]