The distance formula is \(\displaystyle ^{\sqrt{(x_{1}-x_{2})^{2}+(y_{1}-y_{2})^{2}}}\). In the graph shown above the coordinates are \(\displaystyle (-5,8)\) and \(\displaystyle (2,10)\). When you plug the coordinates into the equation you get:

\(\displaystyle \sqrt{(-5-2)^{2}+(8-10)^{2}}\), which then simplifies to \(\displaystyle \sqrt{(-7)^{2}+(2)^{2}}\)

\(\displaystyle \sqrt{49+4}=\sqrt{53}\), because \(\displaystyle 53\) is a prime number there is no need to simplify. 

First, we want to find the value of the diameter of the circle with the given endpoints. We can use the distance formula here:

\(\displaystyle d=\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}=\sqrt{(4-0)^2 + (5-0)^2}=\sqrt{16 +25}=\sqrt{41}\)

If the diameter is \(\displaystyle \sqrt{41}\), then the radius is half of that, or \(\displaystyle \frac{\sqrt{41}}{2}\).

We can then plug that radius value into the formula for the area of a circle.

\(\displaystyle A=\pi r^2 = \pi \cdot (\frac{\sqrt{41}}{2})^2 = \frac{41\pi}{4}\). 

The distance between two points \(\displaystyle (x_1, y_1)\) and \(\displaystyle (x_2, y_2)\) is given by the Distance Formula:

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

Let \(\displaystyle (x_1, y_1) = (0,0)\) and \(\displaystyle (x_2, y_2) = (\pi, \frac{\pi}{2})\). Substitute these values into the Distance Formula.

\(\displaystyle d=\sqrt{(\pi - 0)^2+(\frac{\pi}{2} - 0)^2} = \sqrt{\pi^2 + (\frac{\pi}{2})^2}=\sqrt{\pi^2 + \frac{\pi^2}{4}}\)

To simplify this square root, find a common denominator between the two terms.

\(\displaystyle \sqrt{\pi^2 + \frac{\pi^2}{4}}=\sqrt{\frac{4\pi^2}{4} + \frac{\pi^2}{4}}=\sqrt{\frac{5\pi^2}{4}}\)

Both 4 and \(\displaystyle \pi^2\) are perfect squares, so we can take their square roots to find

\(\displaystyle \sqrt{\frac{5\pi^2}{4}}=\frac{\pi}{2}\sqrt{5}\)

The distance between our two points is \(\displaystyle \frac{\pi}{2}\sqrt{5}\).

The distance between two points \(\displaystyle (x_1, y_1)\) and \(\displaystyle (x_2, y_2)\) is given by the following formula:

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

Let \(\displaystyle (x_1, y_1) = (3, 5)\) and let \(\displaystyle (x_2, y_2) = (9, 11)\). When we plug these two coordinates into the equation we get:

\(\displaystyle d=\sqrt{(9 - 3)^2 + (11 - 5)^2} = \sqrt{6^2 + 6^2} = \sqrt{72} = \sqrt{2 \cdot 36} = \sqrt{2\cdot 6^2} = 6\sqrt{2}\)

Remember that the general equation of a circle with center \(\displaystyle (h, k)\) and radius \(\displaystyle r\) is \(\displaystyle (x-h)^2 + (y-k)^2 = r^2\). 

With this in mind, the center of our circle is \(\displaystyle (2, 3)\). To find the distance from this point to \(\displaystyle (7, 9)\), we can use the distance formula. 

\(\displaystyle d= \sqrt{(x_2-x_1)^2 + (y_2-y_2)^2} = d= \sqrt{(7-2)^2 + (9-3)^2} = \sqrt{5^2 + 6^2} = \sqrt{25+36} = \sqrt{61}\)

To solve this problem let's choose two vertices that lie diagonally from one another. Let's choose \(\displaystyle (3, 1)\) and \(\displaystyle (7, 7)\). 

We can plug these two points into the Distance Formula, and that will give us the length of the box's diagonal. 

\(\displaystyle d= \sqrt{(x_2-x_1)^2 + (y_2-y_2)^2} = \sqrt{(7-3)^2 + (7-1)^2} = \sqrt{4^2 + 6^2} = \sqrt{24+36} = \sqrt{60} = \sqrt{2^2 \cdot 15} = 2\sqrt{15}\)

Step 1: We need to recall the distance formula, which helps us calculate the length of a line between the two points.

The formula is: \(\displaystyle d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\), where \(\displaystyle d=\)distance and \(\displaystyle (x_1,y_1);(x_2,y_2)\) are my two points.

Step 2: We need to identify \(\displaystyle x_1,x_2,y_1,y_2\).

\(\displaystyle x_1=6, x_2=-7, y_1=3, y_2=4\)

Step 3: Substitute the values in step 2 into the formula:

\(\displaystyle \sqrt{(-7-6)^2+(4-3)^2}\)

Step 4: Start evaluating the parentheses:

\(\displaystyle \sqrt {(-13)^2+1^2}\)

Step 5: Evaluate the exponents inside the square root

\(\displaystyle \sqrt{169+1}\)

Step 6: Add the inside:

\(\displaystyle \sqrt{170}\)

Step 7: We need to evaluate \(\displaystyle \sqrt{170}\) in a calculator

\(\displaystyle \sqrt{170}\approx 13.038\approx13.04\)

The length of a segment with endpoints \(\displaystyle (x_{1}, y_{1})\) and \(\displaystyle (x_{2}, y_{2})\) can be calculated using the distance formula:

\(\displaystyle d = \sqrt{(x_{2}- x_{1})^{2} + (y_{2}- y_{1})^{2} }\)

Setting \(\displaystyle x_{2} = y_{2} = N\) and \(\displaystyle x_{1} = y_{1} = \frac{1}{2}\) and substituting:

\(\displaystyle d = \sqrt{ \left (N- \frac{1}{2}\right ) ^{2} + \left (N- \frac{1}{2}\right ) ^{2}}\)

The binomials can be rewritten using the perfect square trinomial pattern:

\(\displaystyle d = \sqrt{ \left [N^{2} - 2 \cdot N \cdot \frac{1}{2} + \left ( \frac{1}{2}\right ) ^{2} \right ] + \left [N^{2} - 2 \cdot N \cdot \frac{1}{2} + \left ( \frac{1}{2}\right ) ^{2} \right ] }\)

\(\displaystyle = \sqrt{\left ( N^{2} - N + \frac{1}{4} \right ) +\left ( N^{2} - N + \frac{1}{4} \right ) }\)

Simplify and collect like terms:

\(\displaystyle d = \sqrt{ N^{2} - N + \frac{1}{4}+ N^{2} - N + \frac{1}{4} }\)

\(\displaystyle = \sqrt{ N^{2} + N^{2} - N- N + \frac{1}{4} + \frac{1}{4} }\)

\(\displaystyle = \sqrt{ 2 N^{2} - 2 N + \frac{1}{2} }\)

The length of a segment with endpoints \(\displaystyle (x_{1}, y_{1})\) and \(\displaystyle (x_{2}, y_{2})\) can be calculated using the distance formula:

\(\displaystyle d = \sqrt{(x_{2}- x_{1})^{2} + (y_{2}- y_{1})^{2} }\)

Setting \(\displaystyle x_{1} = y_{2} = -\frac{1}{4}\) and \(\displaystyle x_{2} = y_{1} = N\), and substituting:

\(\displaystyle d = \sqrt{ \left (N - \left (- \frac{1}{4} \right )\right ) ^{2} + \left ( - \frac{1}{4} - N \right ) ^{2}}\)

\(\displaystyle d = \sqrt{ \left (N + \frac{1}{4} \right ) ^{2} + \left ( - N - \frac{1}{4} \right ) ^{2}}\)

The binomials can be rewritten using the perfect square trinomial pattern:

\(\displaystyle d = \sqrt{ \left [N^{2}+ 2 \cdot N \cdot \frac{1}{4} + \left ( \frac{1}{4}\right ) ^{2} \right ] + \left [(-N)^{2} - 2 \cdot (-N) \cdot\left ( - \frac{1}{4}\right ) + \left ( - \frac{1}{4}\right ) ^{2} \right ] }\)

\(\displaystyle d = \sqrt{ \left ( N^{2}+ \frac{1}{2} N + \frac{1}{16} \right ) +\left ( N^{2}+ \frac{1}{2} N + \frac{1}{16} \right ) }\)

Simplify and collect like terms:

\(\displaystyle d = \sqrt{ N^{2}+ N^{2}+ \frac{1}{2} N + \frac{1}{2} N + \frac{1}{16} + \frac{1}{16} }\)

\(\displaystyle = \sqrt{2 N^{2}+ N + \frac{1}{8} }\)

The length of a segment with endpoints \(\displaystyle (x_{1},y_{1}) , (x_{2},y_{2})\) can be calculated using the distance formula 

\(\displaystyle d = \sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2} }\)

Substituting \(\displaystyle x_{1} = \frac{1}{3} ,y_{1} =-\frac{1}{3}, x_{2} = y_{2}= N\), and simplifying"

\(\displaystyle d = \sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2} }\)

\(\displaystyle = \sqrt{\left (N - \frac{1}{3}\right )^{2}+ \left (N - \left ( - \frac{1}{3} \right ) \right )^{2} }\)

\(\displaystyle = \sqrt{\left (N - \frac{1}{3}\right )^{2}+ \left (N + \frac{1}{3} \right ) ^{2} }\)

\(\displaystyle = \sqrt{ \left [ N^{2} - 2 \cdot N \cdot \frac{1}{3} + \left ( \frac{1}{3}\right )^{2} \right ]+ \left [ N ^{2} + 2 \cdot N \cdot \frac{1}{3} + \left ( \frac{1}{3}\right )^{2} \right ] }\)

\(\displaystyle = \sqrt{ \left ( N^{2} - \frac{2}{3}N + \frac{1}{9} \right ) + \left ( N^{2} + \frac{2}{3}N + \frac{1}{9} \right ) }\)

\(\displaystyle = \sqrt{ N^{2} + N^{2} - \frac{2}{3}N + \frac{2}{3}N + \frac{1}{9} + \frac{1}{9} }\)

\(\displaystyle = \sqrt{ 2N^{2} +\frac{2} {9} }\)