Question

The distance between the points $(3,1)$ and $(0,x)$ is $5$ units. Find x.

Answer 1

We know that the distance between the two points $(x_1,y_1)$ and $(x_2,y_2)$ is

$d=\sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2}$

Here, the given points are $(3,1)$ and $(0,x)$ and the distance is $5$ units, therefore,

$d=\sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2}\Rightarrow 5=\sqrt{{(0-3)}^2+{(x-1)}^2}\Rightarrow 5=\sqrt{{(-3)}^2+{(x-1)}^2}\Rightarrow 5=\sqrt{9+{(x-1)}^2}\Rightarrow 5^2=(\sqrt{9+{(x-1)}^2}{)}^2\Rightarrow 25=9+{(x-1)}^2$

$\Rightarrow {(x-1)}^2=25-9\Rightarrow {(x-1)}^2=16\Rightarrow x-1=\pm \sqrt{16}\Rightarrow x-1=\pm 4\Rightarrow x-1=4,x-1=-4\Rightarrow x=4+1,x=-4+1\Rightarrow x=5,x=-3$

Hence, $x=-3,5$.