Perpendicular Segments

[nycmathguy]

See attachment.

Let me see.

To find d_1 & d_2, use the distance formula.
I then must use the distance again to find the length of the hypotenuse. Finally, express my lengths in terms of the Pythagorean Theorem.

Yes?

 

[MathLover1]

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)

hypotenuse= (d1)^2+(d2)^2

hypotenuse= (sqrt(1+m1^2))^2+(sqrt(1+m2^2))^2

hypotenuse=1+m1^2+1+m2^2
hypotenuse=2+m1^2+m2^2.............eq.1


distance between endpoints of hypotenuse is

hypotenuse=sqrt((1-1)^2+(m2-m1)^2)

hypotenuse=sqrt(0+(m2-m1)^2) ............eq.2

from eq.1 and eq.2 we have

2+m1^2+m2^2=sqrt((m2-m1)^2)...square both sides

(2+m1^2+m2^2)^2=(m2-m1)^2.......take sqrt of both sides


2+m1^2+m2^2=m2-m1

2+m1^2+m2^2-m2+m1 =0.....complete squares

(m1 + 1/2)^2 + (m2 - 1/2)^2 + 3/2 = 0..........solve for m1

(m1 + 1/2)^2 =- (m2 - 1/2)^2 - 3/2

m1 + 1/2 =sqrt(- (m2 - 1/2)^2 - 3/2)

m1 + 1/2 =(1/2) sqrt(-4 (m2 - 1) m2 - 7)

m1 =(1/2) sqrt(-4 (m2 - 1) m2 - 7)-1/2

 

[nycmathguy]

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)

hypotenuse= (d1)^2+(d2)^2

hypotenuse= (sqrt(1+m1^2))^2+(sqrt(1+m2^2))^2

hypotenuse=1+m1^2+1+m2^2
hypotenuse=2+m1^2+m2^2.............eq.1


distance between endpoints of hypotenuse is

hypotenuse=sqrt((1-1)^2+(m2-m1)^2)

hypotenuse=sqrt(0+(m2-m1)^2) ............eq.2

from eq.1 and eq.2 we have

2+m1^2+m2^2=sqrt((m2-m1)^2)...square both sides

(2+m1^2+m2^2)^2=(m2-m1)^2.......take sqrt of both sides


2+m1^2+m2^2=m2-m1

2+m1^2+m2^2-m2+m1 =0.....complete squares

(m1 + 1/2)^2 + (m2 - 1/2)^2 + 3/2 = 0..........solve for m1

(m1 + 1/2)^2 =- (m2 - 1/2)^2 - 3/2

m1 + 1/2 =sqrt(- (m2 - 1/2)^2 - 3/2)

m1 + 1/2 =(1/2) sqrt(-4 (m2 - 1) m2 - 7)

m1 =(1/2) sqrt(-4 (m2 - 1) m2 - 7)-1/2

Great math work as always. What about my reasoning for this thread? Am I wrong in my discussion?

 

[MathLover1]

Great math work as always. What about my reasoning for this thread? Am I wrong in my discussion?

your reasoning for this thread is perfect

 

[nycmathguy]

your reasoning for this thread is perfect

Thanks. More questions tomorrow. By the way, I think we are the only two members actively involved here. What do you think? Good site? Not so good?