# Thread: >>Coordinate Geometry: Find The Distance<<

1. ## >>Coordinate Geometry: Find The Distance<<

yes im new so pls excuse if i do something wrong but,

I have an exam in eight hours and im panicking because im really tired.
Essentially, ive forgotten how to calculate distance in C/G (Coordinate Geometry) Puzzled with the first revision question already:

Find the distance between R(-1, 5) and S(2, -1)

the formula is: $d = \sqrt{(X_{B} - X_{A})^2 + (Y_{B} - Y_{A})^2}$

Could someone please walk me through the process of working this out? thankyoou so much!

The booklet says: $RS = \sqrt{(2-(-1))^2 + (-1-5)^2}$

2. Originally Posted by rooobbbb
yes im new so pls excuse if i do something wrong but,

I have an exam in eight hours and im panicking because im really tired.
Essentially, ive forgotten how to calculate distance in C/G (Coordinate Geometry) Puzzled with the first revision question already:

Find the distance between R(-1, 5) and S(2, -1)

the formula is: $d = \sqrt{(X_{B} - X_{A})^2 + (Y_{B} - Y_{A})^2}$

Could someone please walk me through the process of working this out? thankyoou so much!

$
RS=\sqrt{(-1-2)^2+(-1-5)^2}
$

3. ## The distance formula - geometry's great illusion

The distance formula is one of those equations which looks way more complicated than it is.

1) Forget coordinates for a second. What we are really looking at is the Pythagorean theorem:
$a^2 = b^2 + c^2$
Or, rewritten:
$a = \sqrt{b^2 + c^2}$

We use this theorem to find the length of the hypotenuse in a right triangle. In order to use it, we need a length x and y between the two points you have.

2) So, lets get the length x and y. So the question here is, how far apart are the points (-1,5) and (2, -1) in the x and y direction? For x, we look at the x coordinates -1 and 2.
If these points are on a straight line, -1 and 2 are 3 units apart from one another or (2 - (-1)) = 3.
If the y coordinates are on a straight line, 5 and -1 are 6 units apart from one another or (5 - (-1)) = 6.

3) The distances 3 and 6 are our "a" and "b" values in the Pythagorean theorem. If you put all this mumbo-jumbo together you will get the "distance formula" that you found:
$s = \sqrt{a^2 + b^2}
= \sqrt{(2 - (-1))^2 + (5-(-1))^2}$

4) Simplifying, we get:
$s = \sqrt{3^2 + 6^2}
= \sqrt{9 + 36}
= \sqrt{45}$