Solution:
Given that the point P lies 1/3 along the segment RS as shown below:
To find the y coordinate of the point P, since the point P lies on 1/3 along the segment RS, we have
[tex]\begin{gathered} RP:PS \\ \Rightarrow\frac{1}{3}:\frac{2}{3} \\ thus,\text{ we have} \\ 1:2 \end{gathered}[/tex]Using the section formula expressed as
[tex][\frac{mx_2+nx_1}{m+n},\frac{my_2+ny_1}{m+n}][/tex]In this case,
[tex]\begin{gathered} m=1 \\ n=2 \end{gathered}[/tex]where
[tex]\begin{gathered} x_1=-7 \\ y_1=-2 \\ x_2=2 \\ y_2=4 \end{gathered}[/tex]Thus, by substitution, we have
[tex]\begin{gathered} [\frac{1(2)+2(-7)}{1+2},\frac{1(4)+2(-2)}{1+2}] \\ \Rightarrow[\frac{2-14}{3},\frac{4-4}{3}] \\ =[-4,\text{ 0\rbrack} \end{gathered}[/tex]Hence, the y-coordinate of the point P is
[tex]0[/tex]
