Question

Find the values of x for which the distance between the points P(x, 4) and Q(9, 10) is 10 units.

Answer 1

The given points are P(x, 4) and Q(9, 10).
$$\begin{gathered} \therefore PQ=\sqrt{{\left(x-9\right)}^2+{\left(4-10\right)}^2} \\ =\sqrt{{\left(x-9\right)}^2+{\left(-6\right)}^2} \\ =\sqrt{x^2-18x+81+36} \\ =\sqrt{x^2-18x+117} \end{gathered}$$
$$\begin{gathered} \because PQ=10 \\ \therefore \sqrt{x^2-18x+117}=10 \\ \Rightarrow x^2-18x+117=100\left(\mathrm{Squaring}\mathrm{both}\mathrm{sides}\right) \\ \Rightarrow x^2-18x+17= \end{gathered}$$
$$\begin{gathered} \Rightarrow x^2-17x-x+17=0 \\ \Rightarrow x\left(x-17\right)-1\left(x-17\right)=0 \\ \Rightarrow \left(x-17\right)\left(x-1\right)=0 \\ \Rightarrow x-17=0\mathrm{or}x-1=0 \end{gathered}$$
$\Rightarrow x=17\mathrm{or}x=1$
Hence, the values of x are 1 and 17.