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Find the value of $x$ such that $PQ = QR$ where the coordinates of $P, Q$ and $R$ are $(6, -1), (1, 3)$ and $(x, 8)$ respectively.
Given:
The coordinates of $P, Q$ and $R$ are $(6, -1), (1, 3)$ and $(x, 8)$ respectively.
$PQ = QR$
To do:
We have to find the value of $x$.
Solution:
We know that,
The distance between two points \( \mathrm{A}\left(x_{1}, y_{1}\right) \) and \( \mathrm{B}\left(x_{2}, y_{2}\right) \) is \( \sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}} \).
Therefore,
\( P Q=\sqrt{(1-6)^{2}+(3+1)^{2}} \)
Squaring on both sides, we get,
\( P Q^{2}=(1-6)^{2}+(3+1)^{2} \)
\( =(-5)^{2}+(4)^{2} \)
\( =25+16 \)
\( =41 \)
Similarly,
\( Q R=\sqrt{(x-1)^{2}+(8-3)^{2}} \)
Squaring on both sides, we get,
\( Q R^{2}=(x-1)^{2}+(8-3)^{2} \)
\( =(x-1)^{2}+(5)^{2} \)
\( =(x-1)^{2}+25 \)
\( P Q=Q R \)
\( \Rightarrow 41=(x-1)^{2}+25 \)
\( \Rightarrow(x-1)^{2}=41-25 \)
\( =16 \)
\( =(4)^{2} \)
\( x-1=\pm 4 \)
If \( x-1=4 \) then \( x=1+4=5 \)
If \( x-1=-4 \) then \( x=-4+1=-3 \)
The values of $x$ are $-3$ and $5$.