# Thread: Coordinate Geometry

1. ## Coordinate Geometry

I'm sorry my semi circle isn't drawn very well, but it is meant to be a semi circle.

this is what I did, could you please tell me where I went wrong?

Because the distance from the point of (0,0) to both the bottom right and bottom left point is 10 units, so I assumed that the distance from that point to any point on the semi circle would be 10 units.

From here, I used the distance formula (d=square root of (x2-x1)^2+(y2-y1)^2) putting in (0-s)^2+(0-8)^2 and putting in "10" as d.

So:
10=SQUAREROOT[(0-s)^2+(0-8)^2]

From here I solved until I got d=-2.

But the answer in the back of the book is "6".

Could you please tell me where I went wrong in the process and how to fix it?

If I made a mistake in solving for d and I did everything else right can you show me how to accurately solve for d?

Thank you very much!

2. Are you looking for the distance from the origin (0,0) to a point at (s,8) which lies on the semi-circle

By, the looks of it this seem somewhat to what you are doing.

Recall that the formula of a Circle is $x^2 + y^2 = r^2$

If the distance from the middle of the circle to the end is 10 then this is your Radius or R
and you are already give your Y value

plug in everything and solve

3. Originally Posted by positiveion

I'm sorry my semi circle isn't drawn very well, but it is meant to be a semi circle.

this is what I did, could you please tell me where I went wrong?

Because the distance from the point of (0,0) to both the bottom right and bottom left point is 10 units, so I assumed that the distance from that point to any point on the semi circle would be 10 units.

From here, I used the distance formula (d=square root of (x2-x1)^2+(y2-y1)^2) putting in (0-s)^2+(0-8)^2 and putting in "10" as d.

So:
10=SQUAREROOT[(0-s)^2+(0-8)^2]

From here I solved until I got d=-2.

But the answer in the back of the book is "6".

Could you please tell me where I went wrong in the process and how to fix it?

If I made a mistake in solving for d and I did everything else right can you show me how to accurately solve for d?

Thank you very much!
draw a vertical line from the point (s,8) to the x-axis. do you see that you have a right-triangle, with base length s (which is what you want to find), height 8 and hypotenuse 10. you can use Pythagoras' theorem to find s, which will be 6. or notice that this is a 3-4-5 triangle, all the sides are multiplied by 2. so the base has to be 2*3 = 6, since we have the hypotenuse is 2*5 and the height is 2*4

4. I'm supposed to find what "s" is.

5. Originally Posted by positiveion

this is what I did, could you please tell me where I went wrong?

Because the distance from the point of (0,0) to both the bottom right and bottom left point is 10 units, so I assumed that the distance from that point to any point on the semi circle would be 10 units.

From here, I used the distance formula (d=square root of (x2-x1)^2+(y2-y1)^2) putting in (0-s)^2+(0-8)^2 and putting in "10" as d.

So:
10=SQUAREROOT[(0-s)^2+(0-8)^2]

From here I solved until I got d=-2.

...
1. Your equation is OK.

2. There isn't any "d" in your equation, so where did you get this value from?

3.
$10=\sqrt{(0-s)^2+(0-8)^2}~\implies~ 10 = \sqrt{s^2+64}
$

Now square both sides:

$100 = s^2 +64~\implies~36 = s^2~\implies~s = -6~\vee~\boxed{s=6}$

The negative solution isn't very plausible here.

6. positiveion your approach 10=SQUAREROOT[(0-s)^2+(0-8)^2] is perfect.
You just did some calculation mistake.
If you solve the above thing properly s will come out to be 6 not 2.

Remember you have to find the value of s.