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A triangle $\text{ABC}$ has area $32$ sq units and its side $\text{BC}$, of length $8$ units, lies on the line $x =4$. Then the shortest possible distance between $\text{A}$ and the point $(0,0)$ is

1. $4$ units
2. $8$ units
3. $4\sqrt2$ units
4. $2\sqrt2$ units

Given that, $\triangle ABC$ has area $= 32 \; \text{sq units.}$

Length of $BC = 8 \; \text{units},$ and it lies on the line $x=4.$

Since we want the shortest distance possible between $A$ and the point $(0,0).$

So, we can assume point $A$ lie on the $x\text{-axis}$ and its coordinates be $(p,0).$

Since $|x_{1} – x_{2}|$ is the distance between the $x\text{-coordinates}$ of the two points.

So, the height of the $\triangle ABC:\; \boxed {AD = |p-4| \; \text{units}}$

$\text{The area of triangle} = \frac{1}{2} \times \text{Base} \times \text{Height}$

The area of $\triangle ABC = \frac{1}{2} \times 8 \times |p-4|$

$\Rightarrow 4 \times |p-4| = 32$

$\Rightarrow |p-4| = 8$

$\Rightarrow p-4 = 8 \; (\text{or}) \; – (p-4) = 8$

$\Rightarrow \boxed{ p=12} \; (\text{or}) \; -p+4 = 8 \Rightarrow \boxed{p= -4}$

$\therefore$ The shortest possible distance between $A(p,0) = A(-4,0)$ and the point $O(0,0) = |-4-0| = |-4| = 4 \; \text{units.}$

Correct Answer $: \text{A}$

$\textbf{PS:}$ For any real number $x,$ the absolute value or modulus of $x$ is denoted by $|x|.$

$$|x| = \left\{\begin{matrix} x\;, & \text{if} \; x \geq 0 & \\ -x\;, & \text{if} \; x <0. & \end{matrix}\right.$$

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