(0, b),(3,1); d = 5 Solve for b. I must be stupid or something. I keep getting 15, 17.
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sqr root((3-0)^2)+)(1-b)^2) =5 ((3-0)^2)+)(1-b)^2) = 5^2 1-2b+b^2=9 b^2-2b-8=0 b^2 -4b +2b -8=0 b(b-4)+2(b-4)=0 (b-4)(b+2)=0 b=4,-2.
Are you sure, my calculator keeps giving me 5, -3 Your probably right.
Last edited by Kanderson; Apr 23rd 2012 at 05:09 PM.
No calculator is right,and the one below me is right,i just flipped halfway through the problem,but essentially thats how you do it.
Last edited by syedmaheen; Apr 23rd 2012 at 06:14 PM.
$\displaystyle (3-0)^2 + (1-b)^2 = 5^2$ $\displaystyle 9 + (1 - 2b + b^2) = 25$ $\displaystyle b^2 - 2b - 15 = 0$ $\displaystyle (b - 5)(b + 3) = 0$ $\displaystyle b = 5$ , $\displaystyle b = -3$
Originally Posted by syedmaheen sqr root((3-0)^2)+)(1-b)^2) =5 ((3-0)^2)+)(1-b)^2) = 5^2 So 9+ 1- 2b+ b^2= 25 Subtracting 9 from both sides, 1- 2b+ b^2= 16, not 9. 1-2b+b^2=9 b^2-2b-8=0 b^2 -4b +2b -8=0 b(b-4)+2(b-4)=0 (b-4)(b+2)=0 b=4,-2.
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