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Expert: Scotto - 6/29/2008


Question

notes
Hello,

  If you have a moment I could uses a little assist, I was trying to work this problem from my calc disk and cannot come up with the answer. I know it’s something silly I’m missing but would you please help me out. I’ve been at this thing for three days. Thanks ahead of time. God bless and have a great day.


Determine the value of  " x " that would make the points (x,0), (-2,1) and (3,4) the vertices of a right triangle.

I started with the formula a triangle :

Using the distance formula : D3 = D1 + D2

( ok when I tried to paste the drawings,I had a problem, so I when back in made my notes into an attachment. sorry for the trouble, hope I'm being clear.

   


   



I got 30 = (x^2 +4x +4+1) + x^2 - 6x +9 +16, after this I seem to get lost. The radicands are gone due to squaring both sides of the equation because of the triangle formula. Please let me know what my next step is or where I blew it. I have been banging my head for three days trying to answer this problem. I know the answer is ( -1, 2) For the value of “x”. I just can’t seem to get there.





Later,

  William


Answer
D1²=(x+2)²+(1-0)²=x²+4x+5,
D2²=(x-3)²+(4-0)²=x²-6x+25, and
D3²=(3+2)²+(4-1)²=34.

D1 hypoteneuse
Is it clear which one is the hypoteneuse?  Let's try D1, so we have x²+4x+5 = x²-6x+25 + 34.  Subtract x² from both sides and combine constants giving 4x+5=-6x+59.  This gives 10x=54, or x=5.4.

D2 hypoteneuse
Let's try D2, so we have x²-6x+25 = x²+4x+5 +34.  Again, the x² cancel, so that we're left with -10x=14.  Since we are talking distance, this makes no sense.

D3 hypoteneuse
Let's try D3, so we have x²-6x+25 + x²+4x+5 = 34.  This reduces to 2x²-2x-4=0, or x²-x-2=0.  This factors into (x-2)(x+1).  The distance x can't be negative, so x=2 is the case here.

The solutions were stated to be -1 and 2, but the -1 makes no sense with distances.

Note also that the first equation would work if D1 were the hypoteneuse.

Scotto

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