we have that
JK=JP+PK -----> by the three points are collinear
and
JP/JK=1/3 -----> JK=3JP -----> equation A
substitute
3JP=JP+PK
2JP=PK ------> equation B
Find out the distance JK
we have
J(-4,11) and K(8,-1)
substitute in the formula of distance between two points
[tex]JK=\sqrt[\square]{(-1-11)^2+(8+4)^2}[/tex][tex]JK=12\sqrt[\square]{2^{}}[/tex]
equation A
JK=3JP
[tex]\begin{gathered} 12\sqrt[\square]{2}=3JP \\ JP=4\sqrt[\square]{2} \end{gathered}[/tex]
we have the distance JP
substitute in the formula of distance
[tex]4\sqrt[\square]{2}=\sqrt[\square]{(x+4)^2+(y-11)^2}[/tex]
squared both sides
[tex]32=(x+4)^2+(y-11)^2\text{ ----}\longrightarrow\text{ equation C}[/tex]
equation B
2JP=PK
[tex]2(4\sqrt[\square]{2})=\sqrt[\square]{(x-8)^2+(y+1)^2}[/tex]
squared both sides
[tex]128=(x-8)^2+(y+1)^2\text{ ---}\longrightarrow\text{ equation D}[/tex]
Solve the system of equations C and D
the graphs are circles
the solution is the intersection points both graphs
see the attached figure to better understand the problem
the solution is the point (0,7)
the coordinates of P are (0,7)
