SOLUTION: Find coordinates for the point equidistant from (2,1) (2,-4) (-3,1) Please i really need your help ! thankyou

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Question 442382: Find coordinates for the point equidistant from (2,1) (2,-4) (-3,1)
Please i really need your help ! thankyou

Found 3 solutions by MathLover1, Edwin McCravy, josmiceli:
Answer by MathLover1(20849) About Me  (Show Source):
You can put this solution on YOUR website!

Find coordinates for the point equidistant from (2,1) (2,-4) (-3,1)

Let P(x, y) be the point.
The distance between (x, y) and (2, 1) is:
D+=+sqrt%28%28x+-+2%29%5E2+%2B+%28y+-1%29%5E2%29
The distance between (x, y) and (2, -4) is:
D+=+sqrt%28%28x+-+2%29%5E2+%2B+%28y+%2B4%29%5E2%29

The distance between (x, y) and (-3, 1) is:
D+=+sqrt%28%28x+%2B3%29%5E2+%2B+%28y+-1%29%5E2%29
So we have the following system of equations in terms of x, y, and D:
D+=+sqrt%28%28x+-+2%29%5E2+%2B+%28y+-1%29%5E2%29
D+=+sqrt%28%28x+-+2%29%5E2+%2B+%28y+%2B4%29%5E2%29
D+=+sqrt%28%28x+%2B3%29%5E2+%2B+%28y+-1%29%5E2%29

Square through the square roots:
D%5E2=+%28x+-+2%29%5E2+%2B+%28y+-1%29%5E2
D%5E2+=+%28x+-+2%29%5E2+%2B+%28y+%2B4%29%5E2
D%5E2+=+%28x+%2B3%29%5E2+%2B+%28y+-1%29%5E2

Since these all equal D^2, we have two equations set equal to each other and we will produce a system of two equations that we are all used to.
Setting equation 1 = equation 2:
+%28x+-+2%29%5E2+%2B+%28y+-1%29%5E2=%28x+-+2%29%5E2+%2B+%28y+%2B4%29%5E2


++%28y+-1%29%5E2=+%28y+%2B4%29%5E2
++y+%5E2-2y%2B12=+y%5E2%2B8y+%2B16
++cross%28y+%5E2%29-2y%2B12=+cross%28y%5E2%29%2B8y+%2B16
++-2y%2B12=+8y+%2B16
++-16%2B12=+8y+%2B2y
++-4=+10y

++-4%2F10=+y
++-2%2F5=+y...now find x

Setting equation 1 = equation 3:
+%28x+-+2%29%5E2+%2B+%28y+-1%29%5E2=%28x+%2B3%29%5E2+%2B+%28y+-1%29%5E2

+%28x+-+2%29%5E2+=%28x+%2B3%29%5E2+
+x%5E2-4x+%2B4+=+x%5E2%2B6x+%2B9+
+cross%28x%5E2%29-4x+%2B4+=cross%28+x%5E2%29%2B6x+%2B9+
+-4x+%2B4+=+6x+%2B9+

+-9+%2B4+=+6x+%2B4x+
+-5=+10x+

+-5%2F10=+x+
+-1%2F2=+x+
so, the point is (-1/2, -2/5)

Answer by Edwin McCravy(20054) About Me  (Show Source):
You can put this solution on YOUR website!
Find coordinates for the point equidistant from (2,1) (2,-4) (-3,1)

You don't need to do all that calculating the other tutor did.

First we plot the points:

 

Then we draw the triangle:



That is an isosceles right triangle, so the midpoint of
its hypotenuse is equidistant from the three vertices.

So the midpoint is found by averaging the coordinates
of the endpoints of the hypotenuse:




 
If you like you can draw a circle with that point as the center
and the three given points will be equidistance from the center:





Edwin



Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
Call the point equidistant from the 3 given points (x,y)
The distance, d, from (2,1)
(1) +d%5E2+=+%28+y+-+1+%29%5E2+%2B+%28x+-+2+%29%5E2+
The distance, d, from (2,-4)
(2) +d%5E2+=+%28+y+-+%28-4%29+%29%5E2+%2B+%28x+-+2+%29%5E2+
The distance, d, from (-3,1)
(3) +d%5E2+=+%28+y+-+1+%29%5E2+%2B+%28x+-+%28-3%29+%29%5E2+
-----------------------------
From (1) to (2)
++%28+y+-+1+%29%5E2+%2B+%28x+-+2+%29%5E2+=+%28+y+-+%28-4%29+%29%5E2+%2B+%28x+-+2+%29%5E2+
++%28+y+-+1+%29%5E2+%2B+%28x+-+2+%29%5E2+=+%28+y+%2B+4+%29%5E2+%2B+%28x+-+2+%29%5E2+
Subtract +%28+x+-+2+%29%5E2+ from both sides
++%28+y+-+1+%29%5E2++=+%28+y+%2B+4+%29%5E2+
+y%5E2+-+2y+%2B+1+=+y%5E2+%2B+8y+%2B+16+
+-2y+%2B+1+=+8y+%2B+16+
+10y+=+-15+
+y+=+-3%2F2+
-------------
From (2) to (3)

++%28+y+%2B+4+%29%5E2+%2B+%28x+-+2+%29%5E2+=+%28+y+-+1+%29%5E2+%2B+%28x+%2B+3+%29%5E2+

++%28+5%2F2+%29%5E2+%2B+%28x+-+2+%29%5E2+=+%28+-5%2F2+%29%5E2+%2B+%28x+%2B+3+%29%5E2+
+++%28x+-+2+%29%5E2+=++%28x+%2B+3+%29%5E2+
++x%5E2+-+4x+%2B+4+=+x%5E2+%2B+6x+%2B+9+
+-4x+%2B+4+=+6x+%2B+9+
+10x+=+-5+
+x+=+-1%2F2+
-----------
The coordinates of the point are (-1/2,-3/2)
-----------
check answer:
(1) +d%5E2+=+%28+y+-+1+%29%5E2+%2B+%28x+-+2+%29%5E2+
(1) +d%5E2+=+%28+-3%2F2+-+1+%29%5E2+%2B+%28-1%2F2+-+2+%29%5E2+
(1) +d%5E2+=+%28+-5%2F2+%29%5E2+%2B+%28-5%2F2+%29%5E2+
(1) +d%5E2+=+2%2A%2825%2F4%29+
(1) +d%5E2+=+25%2F2+
(1) +d+=+%284%2Asqrt%282%29%29%2F2+
---------------------
(2) +d%5E2+=+%28+y+-+%28-4%29+%29%5E2+%2B+%28x+-+2+%29%5E2+
(2) +d%5E2+=+%28+-3%2F2+-+%28-4%29+%29%5E2+%2B+%28-1%2F2+-+2+%29%5E2+
(2) +d%5E2+=+%28+5%2F2+%29%5E2+%2B+%28+-5%2F2+%29%5E2+
(2) +d%5E2+=+2%2A%2825%2F4%29+
(2) +d%5E2+=+25%2F2+
(2) +d+=+%284%2Asqrt%282%29%29%2F2+
---------------------
(3) +d%5E2+=+%28+y+-+1+%29%5E2+%2B+%28x+-+%28-3%29+%29%5E2+
(3) +d%5E2+=+%28+-3%2F2+-+1+%29%5E2+%2B+%28-1%2F2+-+%28-3%29+%29%5E2+
(3) +d%5E2+=+%28+-5%2F2+%29%5E2+%2B+%285%2F2+%29%5E2+
(3) +d%5E2+=+2%2A%2825%2F4%29+
(3) +d%5E2+=+25%2F2+
(3) +d+=+%284%2Asqrt%282%29%29%2F2+