**Question:**

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.

**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 *P*(6*,**−*1)*, Q*(1*,*3)* *and *R*(*x,*8).

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

$P Q=\sqrt{(6-1)^{2}+(-1-3)^{2}}$

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

$=\sqrt{25+16}$

$P Q=\sqrt{41}$

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

$Q R=\sqrt{(1-x)^{2}+(3-8)^{2}}$

$Q R=\sqrt{(1-x)^{2}+(-5)^{2}}$

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

$P Q=Q R$

$\sqrt{41}=\sqrt{(1-x)^{2}+(-5)^{2}}$

Squaring on both sides of the equation we get,

$41=(1-x)^{2}+(-5)^{2}$

$41=1+x^{2}-2 x+25$

$15=x^{2}-2 x$

Now we have a quadratic equation. Solving for the roots of the equation we have,

$x^{2}-2 x-15=0$

$x^{2}-5 x+3 x-15=0$

$x(x-5)+3(x-5)=0$

$(x-5)(x+3)=0$

Thus the roots of the above equation are 5 and −3.

Hence the values of ‘*x*’ are.

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