# Coordinate Geometry

• Mar 2nd 2008, 08:41 AM
Danielisew
Coordinate Geometry
Hey, how do you know in a random question with 2 coordinates which is the x1 and y1 and which is the x2 and y2? I keep doing the x1/x2 and y1/y2 the wrong way round!

E.G. The line MN is the diameter of a circle, where M and N are (6,-4) and (0,-2) respectively. Find the radius of the circle

I know the forumla being the square root of (x2-x1)^2 + (y2-y1)^2...

But how on earth do you know from the question which is y2 and y1? I keep getting it wrong!

Thanks
• Mar 2nd 2008, 08:44 AM
Jhevon
Quote:

Originally Posted by Danielisew
Hey, how do you know in a random question with 2 coordinates which is the x1 and y1 and which is the x2 and y2? I keep doing the x1/x2 and y1/y2 the wrong way round!

E.G. The line MN is the diameter of a circle, where M and N are (6,-4) and (0,-2) respectively. Find the radius of the circle

I know the forumla being the square root of (x2-x1)^2 + (y2-y1)^2...

But how on earth do you know from the question which is y2 and y1? I keep getting it wrong!

Thanks

it really doesn't matter. just pick one point to be $(x_1,y_1)$ and the other to be $(x_2,y_2)$. you'll get the same answer either way

(you can try this, do it both ways and you will see. can you tell why this works out?)
• Mar 2nd 2008, 08:48 AM
Danielisew
Well..
Well I just tried it with the following:

X2: -2
X1: -4
Y2: 0
Y1: 6

I end up with the diameter being the square root of 40. Therefore, the radius is the square root of 20? It says the answer is the square root of 10! (Worried)

Thanks for your help, but I am still stuck..
• Mar 2nd 2008, 08:54 AM
Jhevon
Quote:

Originally Posted by Danielisew
Well I just tried it with the following:

X2: -2
X1: -4
Y2: 0
Y1: 6

I end up with the diameter being the square root of 40. Therefore, the radius is the square root of 20? It says the answer is the square root of 10! (Worried)

Thanks for your help, but I am still stuck..

ok, here's how coordinates work. the first value in the pair are the x-values. so x1 and x2 are NOT -4 and -2, respectively. but technically, it also does not matter.

$\underbrace {(6,-4)}_{(x_1,y_1)}$ and $\underbrace {(0,-2)}_{(x_2,y_2)}$

thus, $D = \sqrt{(0 - 6)^2 + (-2 - (-4))^2} = \sqrt{36 + 4} = \sqrt{40} = 2\sqrt{10}$

the radius is half of this, thus it is $\sqrt{10}$

so that was your mistake. 1/2 of sqrt(40) is NOT sqrt(20)
• Mar 2nd 2008, 09:17 AM
Danielisew
Ok thanks very much..

Can you give me a quick lesson on that kinda thing where you cancel down surds? I can't do it and have no idea how square root of 40= 2 multiplied by square root 10..

It would be much appreciated!

Thanks (Talking)
• Mar 2nd 2008, 09:22 AM
Jhevon
Quote:

Originally Posted by Danielisew
Ok thanks very much..

Can you give me a quick lesson on that kinda thing where you cancel down surds? I can't do it and have no idea how square root of 40= 2 multiplied by square root 10..

It would be much appreciated!

Thanks (Talking)

well, this hinges on the fact that we can write roots as powers.

we also know that $(ab)^n = a^n b^n$

so, if we have $\sqrt{ab}$, that is the same as $(ab)^{\frac 12} = a^{\frac 12}b^{\frac 12}$

or equivalently, $\sqrt{ab} = \sqrt{a} \sqrt{b}$

so now on to your example. always look to factor out a number for which you know the square root of.

so, $\sqrt{40} = \sqrt{4 \cdot 10} = \sqrt{4} \sqrt{10}$ and we know the square root of four, that's just 2. so we end up with $2 \sqrt{10}$
• Mar 2nd 2008, 10:02 AM
Danielisew
Ok thanks a lot! I am sorted =)