# Thread: Determining the point that is equidistant from three other points

1. ## Determining the point that is equidistant from three other points

Hello all,

I am new here. Questions like this have been asked before but mine is very specific and I cannot find an answer.

I have the following problem to solve:

Determine the point that is equidistant from the points A(-1,7), B(6,6) and C(5,-1).

The text I am using includes answers, and the answer provided is point (52/25), (153/50). I created a point D(x,y) and plugged the numbers into the distance formula to get AD, BD, and CD. I then created systems of equations setting AD=BD and BD=CD to solve for both variables x and y and I get the answer (2,3), which is close to the answer the text provides but not exact.

I am very confused; if the coordinates for D are (2,3) as I have determined, then it seems that AD, BD, and CD all equal 5, so how could (2,3) not be equidistant? Please kindly explain how to arrive at the apparently correct answer of (52/25, 153/50). Have I done this all wrong?

Thank you for taking the time to read and thanks in advance for any assistance.

2. Originally Posted by DumSpiroSpero80
Hello all,

I am new here. Questions like this have been asked before but mine is very specific and I cannot find an answer.

I have the following problem to solve:

Determine the point that is equidistant from the points A(-1,7), B(6,6) and C(5,-1).

The text I am using includes answers, and the answer provided is point (52/25), (153/50). I created a point D(x,y) and plugged the numbers into the distance formula to get AD, BD, and CD. I then created systems of equations setting AD=BD and BD=CD to solve for both variables x and y and I get the answer (2,3), which is close to the answer the text provides but not exact.

I am very confused; if the coordinates for D are (2,3) as I have determined, then it seems that AD, BD, and CD all equal 5, so how could (2,3) not be equidistant? Please kindly explain how to arrive at the apparently correct answer of (52/25, 153/50). Have I done this all wrong?

Thank you for taking the time to read and thanks in advance for any assistance.
You have done the problem, and then you have checked your answer by finding the distance to each point. I agree that the distance from (2,3) to each of the 3 points is 5 (they all involve "3-4-5" triangles). Looks to me like your answer is correct. The answer in the text is close, but not exact.

BTW, I hope you set the squares of the distances equal, and didn't have the bother of square roots.

3. Originally Posted by DumSpiroSpero80
Hello all,

I am new here. Questions like this have been asked before but mine is very specific and I cannot find an answer.

I have the following problem to solve:

Determine the point that is equidistant from the points A(-1,7), B(6,6) and C(5,-1).

The text I am using includes answers, and the answer provided is point (52/25), (153/50). I created a point D(x,y) and plugged the numbers into the distance formula to get AD, BD, and CD. I then created systems of equations setting AD=BD and BD=CD to solve for both variables x and y and I get the answer (2,3), which is close to the answer the text provides but not exact.

I am very confused; if the coordinates for D are (2,3) as I have determined, then it seems that AD, BD, and CD all equal 5, so how could (2,3) not be equidistant? Please kindly explain how to arrive at the apparently correct answer of (52/25, 153/50). Have I done this all wrong?

Thank you for taking the time to read and thanks in advance for any assistance.

D has co-ordinates (u, v).

$AD = \sqrt{(-1 - u)^2 + (7 - v)^2} = \sqrt{1 + 2u + u^2 + 49 - 14v + v^2} = \sqrt{50 + 2u + u^2 - 14v + v^2}$

$BD = \sqrt{(6 - u)^2 + (6 - v)^2} = \sqrt{36 - 12u + u^2 + 36 - 12v + v^2} = \sqrt{72 - 12u + u^2 - 12v + v^2}$

$CD = \sqrt{(5 - u)^2 + (-1 - v)^2} = \sqrt{25 - 10u + u^2 + 1 + 2v + v^2} = \sqrt{26 - 10u + u^2 + 2v + v^2}$

$AD = BD \implies \sqrt{50 + 2u + u^2 - 14v + v^2} = \sqrt{72 - 12u + u^2 - 12v + v^2} \implies {50 + 2u + u^2 - 14v + v^2} = {72 - 12u + u^2 - 12v + v^2} \implies$

$14u = 22 + 2v \implies u = \dfrac{22 + 2v}{14}.$

$AD = CD \implies \sqrt{50 + 2u + u^2 - 14v + v^2} = \sqrt{26 - 10u + u^2 + 2v + v^2} \implies {50 + 2u + u^2 - 14v + v^2} = 26 - 10u + u^2 + 2v + v^2 \implies$

$12u = -24 + 16v \implies = \dfrac{12(22 + 2v)}{14} = - 24 + 16v \implies 12(22 + 2v) = 14(-24 + 16v) \implies$

$264 + 24v = -336 + 224v \implies 200v = 264 + 336 = \implies$

$v = \dfrac{600}{200} = 3 \implies$

$u = \dfrac{22 + 2 * 3}{14} = 2.$

Now let's check

$\sqrt{(-1 - 2)^2 + (7 - 3)^2} = \sqrt{9 + 16} = 5.$

$\sqrt{(6 - 2)^2 + (6 - 3)^2} = \sqrt{16 + 9} = 5.$

$\sqrt{(5 - 2)^2 + (-1 - 3)^2} = \sqrt{9 + 16} = 5.$

Moral. When you have checked your work twice and it disagrees with the answer key, assume the answer key is wrong. Authors, editors, and publishers are human and so err.

4. Another moral:

When you show your work and ask "proper" question/s - you get prompt, courteous and encouraging answer/s.

5. Thank you very much, DrPhil and JeffM. I thought that I had it right but was reluctant to think that the book was wrong.

Subhotosh Khan, sorry, I will be mindful to post 100% of my work in detail next time. I wrongly thought that the explanation I provided of the work I did was sufficient.

6. Originally Posted by DumSpiroSpero80
Thank you very much, DrPhil and JeffM. I thought that I had it right but was reluctant to think that the book was wrong.

Subhotosh Khan, sorry, I will be mindful to post 100% of my work in detail next time. I wrongly thought that the explanation I provided of the work I did was sufficient.
I think you misinterpreted my praise of your post - I did not come to bury your post.

You got answers within 45 minutes - which is very fast.

You made your question very clear - showed sufficient work - an excellent post.

7. Originally Posted by Subhotosh Khan
I think you misinterpreted my praise of your post - I did not come to bury your post.

You got answers within 45 minutes - which is very fast.

You made your question very clear - showed sufficient work - an excellent post.

OK Subhotosh gets three excuses from the corner in the future. An apt allusion to Shakespear, bravo SK.

8. Originally Posted by JeffM
...Shakespear,
Shakespeare
10 minutes in corner...

9. Hello, DumSpiroSpero80!

$\text{Determine the point that is equidistant from }A(\text{-}1,7),\:B(6,6)\text{ and }C(5,\text{-}1).$

The slope of $AB$ is: .$m_{_{AB}} \:=\:\dfrac{6-7}{6-(\text{-}1)} \:=\:\text{-}\dfrac{1}{7}$

$\text{The slope of }BC\text{ is: }\:m_{_{BC}} \:=\:\dfrac{\text{-}1-6}{5-6} \:=\:7$

$\text{We see that }AB \perp BC.$
That is, $ABC$ is a right triangle with angle $B = 90^o$

$\text{Fact: the point equidistant from the vertices of a right triangle}$
. . . . [$\text{is the }midpoint\: o\!f \:the \:hypotenuse.$

$\text{The hypotenuse is }AC,\text{ from }A(\text{-}1,7)\text{ to }C(5,\text{-}1).$
$\text{Its midpoint is: }\,\left(\dfrac{\text{-}1+5}{2},\,\dfrac{7+(\text{-}1)}{2}\right) \:=\: \color{blue}{(2,\,3)}$

10. Originally Posted by Denis
Shakespeare
10 minutes in corner...
Oh, c'mon. Shakespeare, Shakspur, and Shakespear himself signed his name in many different ways. (I have a copy of a will from 200 years ago in which the name of the deceased is spelled two different ways on the same page.)

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