To answer this, you basically use Pythagoras' Theroem, but instead of:
![c = \sqrt{a^(2) + b^(2)}](https://img.qammunity.org/2020/formulas/mathematics/high-school/8vskakz62wj0qjejkyjobjgogvphbxeeyi.png)
it will be :
![distance = \sqrt{(y - y1)^(2) + (x - x1)^(2) }](https://img.qammunity.org/2020/formulas/mathematics/high-school/x89upns6yhqvav1tz32z49zazoymamg6bp.png)
So you are finding the squareroot of the (difference in y coordinates)² plus (difference in x coordinates) ²:
x is the x-coordinate of (0, -1) (so x = 0)
y is the y-coordinte of (0, -1) ( so y = -1)
x1 is the x coordinate of (3, -3) ( so x1 = 3)
y1 is the y coordinate of (3, -3) (so y1 = -3)
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Now, lets find the distance between the two points, by substituting all of this values into the equation at the top:
(substitute in values)
(simplify: note -1 - - 3 = -1 + 3)
(simplify)
(now square the numbers)
(simplify)
![distance = √(13 )](https://img.qammunity.org/2020/formulas/mathematics/high-school/c0iu8oz2uyvvgz0y6hu0cl6vvxarymqhgc.png)
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Answer:
C.
![√(13)](https://img.qammunity.org/2020/formulas/mathematics/middle-school/jfbz4t6k54lr7efsts9ixi2b3z3e48f469.png)