# Thread: Absolute value (similar to triangle ineq.)

1. ## Mid-coordinate value

How can I show that $\displaystyle (0, \frac{1}{2})$ is the equidistant point from the 3 points: $\displaystyle A(2,2)$, $\displaystyle B(2,-1)$ and $\displaystyle C(-2,-1)$. I know that $\displaystyle (0, \frac{1}{2})$ is the midpoint of $\displaystyle AC$ but how can you conclude it's equidistant to $\displaystyle B$? (is it because of the midpoint of AB?)

2. Originally Posted by dannyc
How can I show that $\displaystyle (0, \frac{1}{2})$ is the equidistant point from the 3 points: $\displaystyle A(2,2)$, $\displaystyle B(2,-1)$ and $\displaystyle C(-2,-1)$. I know that $\displaystyle (0, \frac{1}{2})$ is the midpoint of $\displaystyle AC$ but how can you conclude it's equidistant to $\displaystyle B$? (is it because of the midpoint of AB?)
You can't- it's not equidistant from the three points. You can show that by calculating the three distances from that point to A, B, and C. The distance from (0,1/2) to A and B is 3/2 for both but the distance from (0, 1/2) to C is NOT 3/2. To find the point equidistant from three point, average the coordinates of all three points- ((2+ 2+ 1)/3, (2- 1- 1)/3) ("averaging" does not work for more than three points). To show that is the equidistant point, again- calculate the three distances.

3. I'm still confused Ivy..I'm not sure how to proceed? So I should calculate the distances? The actual question is what point is equidistant to the 3 points (we aren't actually supposed to know it's (0, 1/2) but I was working backwards before..)

4. Originally Posted by dannyc
I'm still confused Ivy..I'm not sure how to proceed? So I should calculate the distances? The actual question is what point is equidistant to the 3 points (we aren't actually supposed to know it's (0, 1/2) but I was working backwards before..)
If you show
AB^2 + BC^2 =AC^2,
then the mid point of AC is equidistant from A, B and C.

5. ## mid coordinate value

Originally Posted by dannyc
How can I show that $\displaystyle (0, \frac{1}{2})$ is the equidistant point from the 3 points: $\displaystyle A(2,2)$, $\displaystyle B(2,-1)$ and $\displaystyle C(-2,-1)$. I know that $\displaystyle (0, \frac{1}{2})$ is the midpoint of $\displaystyle AC$ but how can you conclude it's equidistant to $\displaystyle B$? (is it because of the midpoint of AB?)
The three points form a 3-4-5 right triangle and the midpoint of the hypothenuse is equidistant to each point.You prove it

bjh

6. Originally Posted by dannyc
I'm still confused Ivy..I'm not sure how to proceed? So I should calculate the distances? The actual question is what point is equidistant to the 3 points (we aren't actually supposed to know it's (0, 1/2) but I was working backwards before..)
I recommend you go back and read my answer again. I told you exactly how to find the point equidistant from three given points. And it is not a good idea to "work backwards" from a wrong answer.

7. Originally Posted by HallsofIvy
You can't- it's not equidistant from the three points. You can show that by calculating the three distances from that point to A, B, and C. The distance from (0,1/2) to A and B is 3/2 for both but the distance from (0, 1/2) to C is NOT 3/2. To find the point equidistant from three point, average the coordinates of all three points- ((2+ 2+ 1)/3, (2- 1- 1)/3) ("averaging" does not work for more than three points). To show that is the equidistant point, again- calculate the three distances.
Distance between (0,1/2) and A(2,2) is [2^2 + (3/2)^2]^1/2 = 5/2.
Similarly the distances between the point (0,1/2) and B and C are also 5/2.