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 $5$.

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Updated on: 10-Oct-2022

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