To answer this, you basically use Pythagoras' Theroem, but instead of:
[tex]c = \sqrt{a^{2} + b^{2}}[/tex]
it will be :
[tex]distance = \sqrt{(y - y1)^{2} + (x - x1)^{2} }[/tex]
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)
--------------------------------------------------
Now, lets find the distance between the two points, by substituting all of this values into the equation at the top:
[tex]distance = \sqrt{(y - y1)^{2} + (x - x1)^{2} }[/tex] (substitute in values)
[tex]distance = \sqrt{( 0 -3)^{2} + (-1 - -3)^{2} }[/tex] (simplify: note -1 - - 3 = -1 + 3)
[tex]distance = \sqrt{( -3)^{2} + (-1 +3)^{2} }[/tex] (simplify)
[tex]distance = \sqrt{( -3)^{2} + (2)^{2} }[/tex] (now square the numbers)
[tex]distance = \sqrt{9 + 4 }[/tex] (simplify)
[tex]distance = \sqrt{13 }[/tex]
___________________________________________
Answer:
C. [tex]\sqrt{13}[/tex]
