Thread: Cant solve distance formula problem.

(0, b),(3,1); d = 5 Solve for b. I must be stupid or something. I keep getting 15, 17.

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.

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.

,

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.