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Trig: Find x so the distance between (x,2) and (1,5) is √13
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Distance formula!!
How do you set it up?
(x-2)^2 + (2-5)^2 = Root 13 I believe that is it
x-1 **
(1-x)^2+(5-2)^2=13 1+x^2-2x+9=13 1+x^2-2x-4=0 x^2-2x-3=0 solve for x x^2+1x-3x-3=0 x(x+1)-3(x+1)=0 (x+1)(x-3)=0 x=-1 0r 3
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Thank you Dhashni I wasn't sure if i was supposed to put the square root on top of the 13
:)
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