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

Solution:

The given points are P(x, 4) and Q(9, 10).

$\therefore P Q=\sqrt{(x-9)^{2}+(4-10)^{2}}$

$=\sqrt{(x-9)^{2}+(-6)^{2}}$

$=\sqrt{x^{2}-18 x+81+36}$

$=\sqrt{x^{2}-18 x+117}$

$\because P Q=10$

$\therefore \sqrt{x^{2}-18 x+117}=10$

$\Rightarrow x^{2}-18 x+117=100 \quad$ (Squaring both sides)

$\Rightarrow x^{2}-18 x+17=$

$\Rightarrow x^{2}-17 x-x+17=0$

$\Rightarrow x(x-17)-1(x-17)=0$

$\Rightarrow(x-17)(x-1)=0$

$\Rightarrow x-17=0$ or $x-1=0$

$\Rightarrow x=17$ or $x=1$

Hence, the values of x are 1 and 17.