Question

Find the distance between the points (0, –1) and (3, –3).

Answer 1

To answer this, you basically use Pythagoras' Theroem, but instead of:

$c = \sqrt{a^{2} + b^{2}}$

it will be :

$distance = \sqrt{(y - y1)^{2} + (x - x1)^{2} }$

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:

$distance = \sqrt{(y - y1)^{2} + (x - x1)^{2} }$    (substitute in values)

$distance = \sqrt{( 0 -3)^{2} + (-1 - -3)^{2} }$   (simplify: note -1 - - 3 = -1 + 3)

$distance = \sqrt{( -3)^{2} + (-1 +3)^{2} }$     (simplify)

$distance = \sqrt{( -3)^{2} + (2)^{2} }$           (now square the numbers)

$distance = \sqrt{9 + 4 }$                   (simplify)

$distance = \sqrt{13 }$

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Answer:

C.  $\sqrt{13}$

Answer 2

Answer:

C. √13

Step-by-step explanation:

The distance between two points is given by

d =sqrt( (x2-x1)^2 + (y2-y1)^2)

  = sqrt( (3-0)^2 + (-3--1)^2)

  = sqrt( 3^2 + (-3+1)^2)

  = sqrt( 9+(-2)^2)

  = sqrt( 9+4)

  = sqrt(13)