[ATTACH=full]524[/ATTACH] Can you set this up for me?
use distance formula to find d1 and d2 d1=sqrt((1-0)^2)+(m1-0)^2) d1=sqrt(1+m1^2) d2=sqrt((1-0)^2)+(m2-0)^2) d2=sqrt(1+m2^2) in right triangle hypothenuse c is equal to distance between (1,m1) and (1,m2) c=sqrt((1-1)^2)+(m1-m2)^2) c=sqrt((m1-m2)^2) c=m1-m2 since c^2=d1^2+d2^2 we have (m1-m2 )^2=(sqrt(1+m1^2))^2+(sqrt(1+m2^2))^2 (m1-m2 )^2=1+m1^2+1+m2^2 m1^2-m1*m2+m2^2=m1^2+m2^2+2.........simplify -m1*m2=2......multiply by -1 m1*m2=-2..........solve for m1 m1=-2/m2-> shows relationship between m1 and m2