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`

Answer 2

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

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( (mx_2+nx_1)/(m+n), (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(  (2\cdot 6-9)/(3), (2\cdot 5-4)/(3)\right) = \left( (3)/(3),(6)/(3) \right) =(1,2)`.