Given QT is perpendciular bisector of PR.
So, PT=TR=6x-2y.
Given, PQ=5y-31, QR=2y+5.
Using Pythagoras theorem in triangle PQT,
[tex]\begin{gathered} \text{Hypotenuse}^2=Base^2+Altitude^2 \\ Altitude^2=\text{Hypotenuse}^2-Base^2 \\ QT^2=PQ^2-PT^2\ldots\ldots.(1) \end{gathered}[/tex]
Using Pythagoras theorem in triangle QRT,
[tex]\begin{gathered} \text{Hypotenuse}^2=Base^2+Altitude^2 \\ Altitude^2=\text{Hypotenuse}^2-Base^2 \\ QT^2=QR^2-TR^2\ldots\ldots.(2) \end{gathered}[/tex]
Equate equations (1) and (2).
[tex]PQ^2-PT^2=QR^2-TR^2[/tex]
Since PT=TR, we can write
[tex]\begin{gathered} PQ^2-PT^2=QR^2-PT^2 \\ PQ^2=QR^2 \\ PQ=QR \\ 5y-31=2y+5 \\ 5y-2y=5+31 \\ 3y=36 \\ y=\frac{36}{3}=12 \end{gathered}[/tex]
Now, put y=12 in PQ=5y-31.
[tex]PQ=5\times12-31=29[/tex]
Since PQ=QR, QR=29.
Given, PS=4x+4, SR=7x-17.
Also, PS=SR. Hence,
[tex]\begin{gathered} 4x+4=7x-17 \\ 4+17=7x-4x \\ 21=3x \\ \frac{21}{3}=x \\ 7=x \end{gathered}[/tex]
Put x=7 and y=12 in PT=6x-2y to find PT.
[tex]PT=6\times7-2\times12=18[/tex]
Hence, PT=18.
Since PT=TR, PR=2PT.
Therefore,
[tex]\begin{gathered} PR=2PT \\ =2\times18=36 \end{gathered}[/tex]
Put x=7 in PS=4x+4 .
[tex]PS=4\times7+4=32[/tex]
Since PS=SR, SR=32.
Therefore,
x=7
y=12
PQ=29
QR=29
PS=32
SR=32
PT=18
PR=36
