Absolute value (similar to triangle ineq.)

**dannyc** · March 23rd 2010, 10:32 PM

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

**HallsofIvy** · March 24th 2010, 03:54 AM

Originally Posted by dannyc

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

**dannyc** · March 24th 2010, 04:56 AM

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..)

**sa-ri-ga-ma** · March 24th 2010, 05:22 AM

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.

**bjhopper** · March 24th 2010, 09:37 AM

Originally Posted by dannyc

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

**HallsofIvy** · March 25th 2010, 02:57 AM

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.

**sa-ri-ga-ma** · March 25th 2010, 06:42 AM

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.

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