If Q (0,1) is equidistant from P (5, −3) and R (x, 6),

Question:

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.

Solution:

The distance $d$ between two points $\left(x_{1}, y_{1}\right)$ and $\left(x_{2}, y_{2}\right)$ is given by the formula

$d=\sqrt{\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}-y_{2}\right)^{2}}$

The three given points are Q(0,1), P(5,3) and R(x,6).

Now let us find the distance between ‘P’ and ‘Q’.

$P Q=\sqrt{(5-0)^{2}+(-3-1)^{2}}$

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

$=\sqrt{25+16}$

$P Q=\sqrt{41}$

Now, let us find the distance between ‘Q’ and ‘R’.

$Q R=\sqrt{(0-x)^{2}+(1-6)^{2}}$

$Q R=\sqrt{(-x)^{2}+(-5)^{2}}$

It is given that both these distances are equal. So, let us equate both the above equations,

$P Q=Q R$

$\sqrt{41}=\sqrt{(-x)^{2}+(-5)^{2}}$

Squaring on both sides of the equation we get,

$41=(-x)^{2}+(-5)^{2}$

 

$41=x^{2}+25$

$x^{2}=16$

$x=\pm 4$

Hence the values of ' $x$ ' are 4 or $-4$.

 

Now, the required individual distances,

$Q R=\sqrt{(0 \mp 4)^{2}+(1-6)^{2}}$

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

$Q R=\sqrt{41}$

Hence the length of ' $Q R$ ' is $\sqrt{41}$ units.

For ‘PR’ there are two cases. First when the value of ‘x’ is 4

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

$=\sqrt{(1)^{2}+(-9)^{2}}$

 

$=\sqrt{1+81}$

$P R=\sqrt{82}$

Then when the value of ‘x’ is −4,

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

$=\sqrt{(9)^{2}+(-9)^{2}}$

 

$=\sqrt{81+81}$

$P R=9 \sqrt{2}$

Hence the length of ' $P R$ can be $\sqrt{82}$ or $9 \sqrt{2}$ units

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