SOLUTION: Missing coordinate A(-3,2) , B(x,3) , c(4,5) if AB=BC , Find the value of x???

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Question 885250: Missing coordinate A(-3,2) , B(x,3) , c(4,5) if AB=BC , Find the value of x???
Found 2 solutions by richwmiller, algebrapro18:
Answer by richwmiller(17219) About Me  (Show Source):
You can put this solution on YOUR website!
(x,3) is not on the same line as A(-3,2) C(4,5) so we cannot just find the midpoint
The distance (denoted by d) between two points in two dimensions is given by the following formula:
d=sqrt%28%28x1-x2%29%5E2+%2B+%28y1-y2%29%5E2%29
sqrt((4-x)^2 + (5-3)^2)=sqrt((-3-x)^2 + (2-3)^2)
x = 5/7

Answer by algebrapro18(249) About Me  (Show Source):
You can put this solution on YOUR website!
To find x we need to find the distance between A and B and the distance between B and C. We can then set the two distances equal to each other and solve for x, since AB=BC.

So finding the distance between A and B we get:

AB=sqrt%28%28x2-x1%29%5E2%2B%28y2-y1%29%5E2%29
AB=sqrt%28%28-3-x%29%5E2%2B%282-3%29%5E2%29
AB=sqrt%28x%5E2%2B6x%2B9%2B%28-1%29%5E2%29
AB=sqrt%28x%5E2%2B6x%2B9%2B1%29
AB=sqrt%28x%5E2%2B6x%2B10%29

That's as simplified as that can get. So now we find the distance between B and C:

BC=sqrt%28%28x2-x1%29%5E2%2B%28y2-y1%29%5E2%29
BC=sqrt%28%28x-4%29%5E2%2B%283-5%29%5E2%29
BC=sqrt%28x%5E2-8x%2B16%2B%28-2%29%5E2%29
BC=sqrt%28x%5E2-8x%2B16%2B4%29
BC=sqrt%28x%5E2-8x%2B20%29

So now we can set AB = BC and solve for x.
sqrt%28x%5E2%2B6x%2B10%29 = sqrt%28x%5E2-8x%2B20%29 square both sides to get rid of the square root
x%5E2%2B6x%2B10+=+x%5E2-8x%2B20 Subtract (x2-8x+20) from both sides
x%5E2-x%5E2%2B6x%2B8x%2B10-20+=+0 combine like terms
14x-10+=+0 add 10 to both sides
14x+=+10 divide both sides by 14
x = 10/14 = 5/7