Here, we need to use the distance formula between the two points (0, 0) and (3, 4).

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

$\displaystyle d=\sqrt{(3-0)^2+(4-0)^2}$

$\displaystyle d=\sqrt{3^2+4^2}=\sqrt{9+16}=\sqrt{25}=5$

To determine how long the line needs to be to connect those two points, we need to use the distance formula, shown below.

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

The two points are $\displaystyle (x_1,y_1)$ and $\displaystyle (x_2,y_2)$.  In our case, the points are (–8, 3) and (6, –1).

$\displaystyle d=\sqrt{(6-(-8))^2+(-1-3)^2}$

$\displaystyle d=\sqrt{14^2+(-4)^2}$

$\displaystyle d=\sqrt{196+14}$

$\displaystyle d=\sqrt{212}$

$\displaystyle d=\sqrt{4(53)}$

$\displaystyle d=2\sqrt{53}$

So in order to connect the two points, the length of the line needs to have $\displaystyle d=2\sqrt{53}$.

To solve problems like this, we simply need to use the distance formula, $\displaystyle d = \sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}$. Plugging in the $\displaystyle x$ and $\displaystyle y$ values from our points yields $\displaystyle d = \sqrt{(10-7)^{2}+(4-8)^{2}}$, or $\displaystyle d = \sqrt{(3)^{2}+(-4)^{2}}$. Solving this radical gives us a value of $\displaystyle d = \sqrt{25}$, or 5.

Use the distance formula, with  $\displaystyle x_{1} =2,x_{2} =7,y_{1} =9,y_{2} =6$ :

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

$\displaystyle = \sqrt{ \left (2 -7\right ) ^{2} + \left (9 -6 \right ) ^{2} }$

$\displaystyle = \sqrt{ \left (-5 \right ) ^{2} + 3 ^{2} } = \sqrt{34} \approx 5.8$

Therefore, none of the integer answer choices are correct.

The distance between two points can be found with the equation $\displaystyle \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$. Substituting in values you get $\displaystyle \sqrt{(2+4)^2+(1-2)^2}$. This means that the answer is $\displaystyle \sqrt{37}$.

Use the midpoint formula:

$\displaystyle \frac{x_{2}+x_{1}}{2}, \frac{y_{2}+y_{1}}{2}$

Remember points are written in the following format:

$\displaystyle (x,y)$

Substitute for line A

$\displaystyle \frac{\left ( -3+3 \right )}{2} , \frac{\left ( 4+0 \right )}{2}$

$\displaystyle \frac{\left ( 0\right )}{2} , \frac{\left ( 4\right )}{2}$

The midpoint of line A is $\displaystyle (0,4)$.

Substitute for line B.

$\displaystyle \frac{\left ( -2+2 \right )}{2} , \frac{\left ( -2 +2 \right )}{2}$

$\displaystyle \frac{\left ( 0 \right )}{2} , \frac{\left ( 0 \right )}{2}$

The midpoint of line B is $\displaystyle (0,0)$.

Now we can find the distance between these two points using the distance formula:

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

Substitute the using the known values for lines A and B.

$\displaystyle d= \sqrt{\left ( 0-0 \right )^{2} + \left ( 0-4 \right )^{2}}$

Simplify.

$\displaystyle d= \sqrt{\left ( 0\right )^{2} + \left ( -4 \right )^{2}}$

$\displaystyle d= \sqrt{\left ( -4 )^{2}}$

$\displaystyle d= \sqrt{16}$

The distance between the two midpoints of lines A and B is $\displaystyle 4$.

Use the equation to calculated the distance between two points: 

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

where

$\displaystyle (3,1)=(x_1,y_1), \: (-2,7)=(x_2,y_2)$

we can find the distance.

$\displaystyle d=\sqrt{(-2-3)^{2}+(7-1)^{2}}$

   $\displaystyle =\sqrt{25+36}$

   $\displaystyle =\sqrt{61}$

It is definately possible to find the distance from point A to point B, given the coordinates.

We can do this by using the formula: 

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

The points provided can be plugged into this formula as follows:

$\displaystyle \\ \small \small \small d=\small \sqrt{(12-2)^{2}+(20-4)^{2}} \\ \\ = \sqrt{10^{2}+16^{2}}\\ \\ = \sqrt{100+256} \\ \\= \sqrt {356}$.

This is the length.  

It is definately possible to find the distance from point A to point B, given the coordinates:

$\displaystyle \small (4,6)(10,2)$. 

We can do this by using the formula: 

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

The points provided can be plugged into this formula as follows: 

$\displaystyle \\ \small \small \small \small d=\small \sqrt{(10-4)^{2}+(2-6)^{2}} \\ \\= \sqrt{6^{2}+(-4)^{2}} \\ \\= \sqrt{36+16} \\ \\= \sqrt {52}$.

This is the length.