Subtract 3 from 5 and 1 from 2. Now use Phythagorus' Theorum (A^2 + B^2 = C^2). 2^2 + 1^2 = X^2 so 4 + 1 = X^2 or 5 = X^2 or X = square root of 5

Ah...this brings back memories in Geometry...

Ok, so when two coordinates are given and you need to find the distance, there is a special formula you use called the Distance Formula.

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

So when you are trying to find the distance, using the given coordinates, you plug in the numbers for where the x and y are supposed to go. X1 represents the x coordinate for the first coordinate pair, and x2 represents the x coordinate for the second coordinate pair. Y1 represents the y coordinate for the first coordinate pair, and y2 represents the y coordinate for the second coordinate pair. In this case, x1 = 3, x2 = 5, y1 = 2, y2 = 1. Now, you plug in the numbers into the distance formula.

D = sqrt((5 - 3)^2 + (1 - 2)^2)

After you have done this, you must now simplify the equation. Add the numbers in between the parenthesis, and then after that, you square each term. So five minus three is two, and two squared is FOUR. One minus two is negative one, and negative one squared is just ONE.

D = sqrt(2^2 + (-1)^2)

= sqrt(4 + 1)

Now, you will have the square root of five. At this point, your teacher might say to leave the answer as a square root, but if he or she does not, then this is your chance to use a calculator.

Sqrt(5) = 2.23606798

Therefore, your answer could be either sqrt(5) or 2.23606798.

I hope I did not confuse you...:p

EDIT: If your teacher says to not leave it as a square root, he or she might say to round it to the nearest tenth most likely...so your answer (non-square root) would be approximately 2.2. Also when you put an equal sign and then this irrational number, your teacher will tell you to put this type of equal sign where the lines of the equal sign are squiggly. Just saying :P)

Ok, so when two coordinates are given and you need to find the distance, there is a special formula you use called the Distance Formula.

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

So when you are trying to find the distance, using the given coordinates, you plug in the numbers for where the x and y are supposed to go. X1 represents the x coordinate for the first coordinate pair, and x2 represents the x coordinate for the second coordinate pair. Y1 represents the y coordinate for the first coordinate pair, and y2 represents the y coordinate for the second coordinate pair. In this case, x1 = 3, x2 = 5, y1 = 2, y2 = 1. Now, you plug in the numbers into the distance formula.

D = sqrt((5 - 3)^2 + (1 - 2)^2)

After you have done this, you must now simplify the equation. Add the numbers in between the parenthesis, and then after that, you square each term. So five minus three is two, and two squared is FOUR. One minus two is negative one, and negative one squared is just ONE.

D = sqrt(2^2 + (-1)^2)

= sqrt(4 + 1)

Now, you will have the square root of five. At this point, your teacher might say to leave the answer as a square root, but if he or she does not, then this is your chance to use a calculator.

Sqrt(5) = 2.23606798

Therefore, your answer could be either sqrt(5) or 2.23606798.

I hope I did not confuse you...:p

EDIT: If your teacher says to not leave it as a square root, he or she might say to round it to the nearest tenth most likely...so your answer (non-square root) would be approximately 2.2. Also when you put an equal sign and then this irrational number, your teacher will tell you to put this type of equal sign where the lines of the equal sign are squiggly. Just saying :P)