Question

Find the value(s) of $x$, if the distance between the points $A(0,0)$ and $B(x,–4)$ is $5$ units.

Answer 1

Given:

The distance between the two points $A(0,0)$ and $B(x,–4)$ is $5$ units.

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

$=\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2}$
$\Rightarrow \sqrt{(x-0{)}^2+(-4-0{)}^2}=5$

$\Rightarrow \sqrt{x^2+16}=5$

$\Rightarrow x^2+16=5^2=25$

$\Rightarrow x^2=25-16$

$\Rightarrow x^2=9$

$\Rightarrow x=\pm 3$

Hence, the value(s) of $x$ are $3$ and ​$–3$.