Question

If the distance between the points $(3,x)$ and $(-2,-6)$ is $13$ units, then find the value of $x$.

Answer 1

Let $A$ and $B$ be the two points such that $A(3,x)$ and $(-2,-6)$ respectively.

$AB=13\text{units}\left(\text{Given}\right)$

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

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

Therefore,

$AB=\sqrt{{(-2-3)}^2+{(-6-x)}^2}$

$\Rightarrow AB=\sqrt{25+(36+x^2+12x)}$

But given that $AB=13\text{units}$

$\therefore \sqrt{25+(36+x^2+12x)}=13$

Squaring both sides, we have

${\left(\sqrt{25+(36+x^2+12x)}\right)}^2={\left(13\right)}^2$

$25+36+x^2+12x=139$

$x^2+12x+61-139=0$

$x^2+12x-78=0$

Now by quadratic formula, we have

$x=\frac{-12\pm \sqrt{{\left(12\right)}^2-4(-78)\left(1\right)}}{2\times 1}$

$\Rightarrow x=\frac{-12\pm \sqrt{144+312}}{2}$

$\Rightarrow x=\frac{-12\pm \sqrt{456}}{2}$

$\Rightarrow x=-6\pm \sqrt{114}$

Hence the value of $x$ is $(-6+\sqrt{114})$ or $(-6-\sqrt{114})$.