Question

a line segment has endpoints s(-9, -4) and t (6,5). Point R lies on ST such that the ratio of SR to RT is 2:1 What are the coordinates of point R

Answer 1

$\bf ~~~~~~~~~~~~\textit{internal division of a line segment} \\\\\\ S(-9,-4)\qquad T(6,5)\qquad \qquad \stackrel{\textit{ratio from S to T}}{2:1} \\\\\\ \cfrac{S\underline{R}}{\underline{R} T} = \cfrac{2}{1}\implies \cfrac{S}{T} = \cfrac{2}{1}\implies 1S=2T\implies 1(-9,-5)=2(6,5)\\\\ -------------------------------$

$\bf R=\left(\cfrac{\textit{sum of "x" values}}{\textit{sum of ratios}}\quad ,\quad \cfrac{\textit{sum of "y" values}}{\textit{sum of ratios}}\right)\\\\ -------------------------------\\\\ R=\left(\cfrac{(1\cdot -9)+(2\cdot 6)}{2+1}\quad ,\quad \cfrac{(1\cdot -5)+(2\cdot 5)}{2+1}\right) \\\\\\ R=\left(\cfrac{-9+12}{3}~~,~~\cfrac{-5+10}{3} \right)\implies R=\left(\cfrac{3}{3}~~,~~\cfrac{5}{3} \right) \\\\\\ R=\left(1~,~1\frac{2}{3} \right)$

Answer 2

Answer:

The coordinates of the point R are (1,2).

Step-by-step explanation:

Notice that we want to complete an internal division of the segment with endpoints $S(x_1,y_1)$ and $T(x_2,y_2)$, in a given proportion m:n. This means, to find a point $R(x,y)$ such that SR/RT = 1/2.

The formula to obtain the coordinates of the point $R(x,y)$ is:

$(x,y) = \left( \frac{mx_2+nx_1}{m+n}, \frac{my_2+ny_1}{m+n} \right)$.

In our particular case $m=2$, $n=1$, $(x_1,y_1) = (-9,-4)$  and $(x_2,y_2)=(6,5)$. Thus,

$(x,y) = \left(  \frac{2\cdot 6-9}{3}, \frac{2\cdot 5-4}{3}\right) = \left( \frac{3}{3},\frac{6}{3} \right) =(1,2)$.