Question

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

Answer 1

It is given that the distance between the points P (x, 4) and Q (9, 10) is 10 units.
$Let{x}_1=x,y_1=4,x_2=9,y_2=10$
Applying distance formula, if is obtained.
$d=\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2}$
$10=\sqrt{(9-x{)}^2+(10-4{)}^2}$
$10=\sqrt{81+x^2-18x+36}$
$10=\sqrt{x^2-18x+117}$
On squaring both sides, it is obtained.
$100=x^2-18x+117$
$\Rightarrow x^2-18x+17=0$
$\Rightarrow x^2-17x-x+17=0$
$\Rightarrow x(x-17)-1(x-17)=0$
$\Rightarrow (x-1)(x-17)=0$
$\Rightarrow x=1,17$
Thus, the values of x are 1 and 17