In a diagram,
On the other hand, the formula to obtain the distance between two points is
[tex]\begin{gathered} A=(x_1,y_1),B=(x_2,y_2) \\ \Rightarrow d(A,B)=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2} \end{gathered}[/tex]Therefore, in our case,
[tex]\begin{gathered} 2*d(P,S)=d(P,R) \\ \end{gathered}[/tex]After setting R=(x,y), we get,
[tex]\begin{gathered} \Rightarrow2*\sqrt{(x-2)^2+(y-4)^2}=\sqrt{(x+7)^2+(y+2)^2} \\ \Rightarrow4((x-2)^2+(y-4)^2)=(x+7)^2+(y+2)^2 \\ \Rightarrow(3x^2-30x-33)+(3y^2-36y+60)=0 \\ \Rightarrow(x^2-10x-11)+(y^2-12y+20)=0 \end{gathered}[/tex]Furthermore, point P lies along RS, and the equation of such segment is
[tex]\begin{gathered} equation\text{ RS} \\ y=\frac{2}{3}x+\frac{8}{3} \end{gathered}[/tex]Substitute into the square root equation, so the expression only depends on x,
[tex]\begin{gathered} \Rightarrow x^2-10x-11+\frac{4}{9}(x^2+8x+16)-8x-32+20=0 \\ \end{gathered}[/tex]Solve for x, as shown below
[tex]\begin{gathered} \Rightarrow\frac{13}{9}(x^2-10x-11)=0 \\ \Rightarrow x=-1,11 \end{gathered}[/tex]Notice that x=11 would imply that P is not between points R and S but to the 'right' of S.
Therefore, the only valid solution is x=-1.
Calculate the corresponding value of y for x=-1 using the equation of the line RS, as shown below
[tex]\begin{gathered} x=-1 \\ \Rightarrow y=\frac{2}{3}(-1)+\frac{8}{3}=\frac{6}{3}=2 \\ \Rightarrow P=(-1,2) \end{gathered}[/tex]
