Find the Distance Between Two Points (6 square root of 3,9) , (-2 square root of 3,8)

$(6 \sqrt{3}, 9)$ , $(- 2 \sqrt{3}, 8)$

Use the distance formula to determine the distance between the two points.

$Distance = \sqrt{{(x_{2} - x_{1})}^{2} + {(y_{2} - y_{1})}^{2}}$

Substitute the actual values of the points into the distance formula.

$\sqrt{{((- 2 \sqrt{3}) - 6 \sqrt{3})}^{2} + {(8 - 9)}^{2}}$

Simplify.

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Subtract $6 \sqrt{3}$ from $- 2 \sqrt{3}$ .

$\sqrt{{(- 8 \sqrt{3})}^{2} + {(8 - 9)}^{2}}$

Simplify the expression.

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Apply the product rule to $- 8 \sqrt{3}$ .

$\sqrt{{(- 8)}^{2} {\sqrt{3}}^{2} + {(8 - 9)}^{2}}$

Raise $- 8$ to the power of $2$ .

$\sqrt{64 {\sqrt{3}}^{2} + {(8 - 9)}^{2}}$

Rewrite ${\sqrt{3}}^{2}$ as $3$ .

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Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{3}$ as $3^{\frac{1}{2}}$ .

$\sqrt{64 {(3^{\frac{1}{2}})}^{2} + {(8 - 9)}^{2}}$

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$\sqrt{64 \cdot 3^{\frac{1}{2} \cdot 2} + {(8 - 9)}^{2}}$

Combine $\frac{1}{2}$ and $2$ .

$\sqrt{64 \cdot 3^{\frac{2}{2}} + {(8 - 9)}^{2}}$

Cancel the common factor of $2$ .

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Cancel the common factor.

$\sqrt{64 \cdot 3^{\frac{2}{2}} + {(8 - 9)}^{2}}$

Divide $1$ by $1$ .

$\sqrt{64 \cdot 3^{1} + {(8 - 9)}^{2}}$

Evaluate the exponent.

$\sqrt{64 \cdot 3 + {(8 - 9)}^{2}}$

Simplify the expression.

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Multiply $64$ by $3$ .

$\sqrt{192 + {(8 - 9)}^{2}}$

Subtract $9$ from $8$ .

$\sqrt{192 + {(- 1)}^{2}}$

Raise $- 1$ to the power of $2$ .

$\sqrt{192 + 1}$

Add $192$ and $1$ .

$\sqrt{193}$

The result can be shown in multiple forms.

Exact Form:

$\sqrt{193}$

Decimal Form:

$13.89244398 \dots$

Find the Distance Between Two Points (6 square root of 3,9) , (-2 square root of 3,8)

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