Respuesta :
to find Q we need to make 2 distance measure
that they fulfill these conditions
PS=RQ and RS=QP
the formula of distances between 2 points is
[tex]\sqrt[]{(x2-x1)^2+(y2-y1)^2}[/tex]Distance PS
[tex]\begin{gathered} \sqrt[]{(6-4)^2+(-1-(-5))^2} \\ \\ PS=\sqrt[]{20} \end{gathered}[/tex]Distance RQ
[tex]\begin{gathered} \sqrt[]{(0-x)^2+(-1-y)^2} \\ \\ RQ=\sqrt[]{x^2+(1+y)^2} \end{gathered}[/tex]where x and y are de coordinates of Q
Distance RS
[tex]\begin{gathered} \sqrt[]{(0-4)^2+(-1-(-5))^2} \\ \\ RS=\sqrt[]{32} \end{gathered}[/tex]Distance QP
[tex]\begin{gathered} \sqrt[]{(x-6)^2+(y-(-1))^2} \\ \\ QP=\sqrt[]{(x-6)^2+(y+1)^2} \end{gathered}[/tex]now solve the equals
PS=RQ
[tex]\begin{gathered} \sqrt[]{20}=\sqrt[]{x^2+(1+y)^2} \\ 20=x^2+(1+y)^2 \end{gathered}[/tex]RS=QP
[tex]\begin{gathered} \sqrt[]{32}=\sqrt[]{(x-6)^2+(y+1)^2} \\ 32=(x-6)^2+(y+1)^2 \end{gathered}[/tex]if I subtract the two equations I will get
[tex]32-20=(x-6)^2-x^2[/tex]and i will solve to find x
[tex]\begin{gathered} 12=-12x+36 \\ 12x=36-12 \\ x=\frac{24}{12} \\ \\ x=2 \end{gathered}[/tex]the value of x is 2, then I can replace x on any equation to find y
so replacing
[tex]\begin{gathered} 20=x^2+(1+y)^2 \\ 20=(2)^2+y^2+2y+1 \\ y^2+2y+1-20+4=0 \\ y^2+2y-15=0 \end{gathered}[/tex]use factor to solve y
[tex]\begin{gathered} y=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a} \\ \\ y=\frac{-2\pm\sqrt[]{4+60}}{2} \\ \\ y=\frac{-2\pm8}{2} \\ \\ y=-1\pm4 \end{gathered}[/tex]then y will have two values
[tex]\begin{gathered} y_1=-1+4=3 \\ y_2=-1-4=-5 \end{gathered}[/tex]the real coordinate is y=3 because if is y=-5 the point dont form a rectangle
if x=2 and y=3 the point Q is (2,3)
