Perpendicular Segments

[nycmathguy]

Can you set this up for me?

 

[MathLover1]

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

 

[nycmathguy]

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

Another nice, informative reply.