Find the value of x such that PQ = QR
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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