If $Q (0, 1)$ is equidistant from $P (5, -3)$ and $R (x, 6)$, find the values of $x$. Also, find the distances $QR$ and $PR$.

Given:

$Q (0, 1)$ is equidistant from $P (5, -3)$ and $R (x, 6)$.

To do:

We have to find the values of $x$ and the distances $QR$ and $PR$.

Solution:

$Q (0, 1)$ is equidistant from $P (5, -3)$ and $R (x,6)$.

This implies,

\( \mathrm{PQ}=\mathrm{RQ} \)

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,

\( \mathrm{PQ}=\sqrt{(5-0)^{2}+(-3-1)^{2}} \)

\( =\sqrt{5^2+4^2} \)

\( \mathrm{RQ}=\sqrt{(x-0)^{2}+(6-1)^{2}} \)

\( =\sqrt{x^2+5^2} \)

This implies,

\( \sqrt{(5)^{2}+(4)^{2}}=\sqrt{(x)^{2}+(5)^{2}} \)

Squaring on both sides, we get,

\( 25+16=x^{2}+25 \)

\( x^{2}=16 \)

\( x^2=(4)^2 \)

\( x=\pm 4 \)

If \( x=4 \), then,

\( QR=\sqrt{4^2+5^2}=\sqrt{16+25}=\sqrt{41} \)

\( P R=\sqrt{(5-4)^{2}+(-3-6)^{2}} \)
\( =\sqrt{(1)^{2}+(-9)^{2}} \)

\( =\sqrt{1+81} \)

\( =\sqrt{82} \)

If \( x=-4, \) then,

\( QR=\sqrt{(-4)^2+5^2}=\sqrt{16+25}=\sqrt{41} \)
\( P R=\sqrt{(5+4)^{2}+(-3-6)^{2}} \)
\( =\sqrt{(9)^{2}+(-9)^{2}} \)

\( =\sqrt{81+81} \)
\( =\sqrt{81 \times 2} \)

\( =9 \sqrt{2} \)

The distance $QR$ is $\sqrt{41}$ and the distance $PR$ is $\sqrt{82}$ or $9\sqrt{2}$.

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Updated on: 10-Oct-2022

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