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

It is known that the distance between the points $\left(x_1,y_1\right),\left(x_2,y_2\right)$
is $\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}$

According to question,

$$\begin{gathered} PQ=10=\sqrt{{\left(x-9\right)}^2+{\left(4-10\right)}^2} \\ \Rightarrow {\left(x-9\right)}^2+36=100 \\ \Rightarrow {\left(x-9\right)}^2=64 \\ \Rightarrow x^2+81-18x-64=0 \\ \Rightarrow x^2-18x+17=0 \\ \Rightarrow x^2-17x-x+17=0 \\ \Rightarrow x\left(x-17\right)-\left(x-17\right)=0 \\ \Rightarrow \left(x-1\right)\left(x-17\right)=0 \\ \Rightarrow x=1,17 \end{gathered}$$

Hence, the required values of x will be 1 and 17.