Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

It's hard to create question names that make sense. Anyhow, the following is another question from my math assignment.

enter image description here

Line-segment AB has a fixed length of 10 units. point A moves on the positive x-axis and point B moves on the positive y-axis.

Point M is in the middle of line AB.

All right, I have been able to figure out the first question of the assignment: that M always stays on a circle from the Origin with a radius of 5.

However, the second question asks me to prove my answer from the first question.

How do I prove that M will always stay on this circle?

share|improve this question
    
This won't help you on your quest for an answer, but when cutting glass you can use a setup like this. You need two perpendicular rails on which $A$ and $B$ can move freely (past the origin too), and you put a cutter at $M$. If you put $M$ somewhere on $AB$ other than the midpoint (doesn't even have to be between $A$ and $B$), then $M$ will follow an ellipse. –  Arthur Nov 17 '12 at 15:43
add comment

3 Answers

up vote 1 down vote accepted

To prove that $M$ always stays on the circle with center $O$, and radius 5, you have to prove that $OM = 5$, no matter where $A, B$ are.

There is a theorem on right triangle, saying that: "The median on the hypotenuse of a right triangle equals one-half the hypotenuse", and remember that $AB = 10$, can you prove $OM = 5$?

share|improve this answer
add comment

Let $M(h,k), A(a,0)$ and $B(0,b)$

So, using Pythagoras Theorem, $a^2+b^2=10^2$

As $M$ is the midpoint, $$h=\frac{a+0}2,k=\frac{0+b}2\implies a=2h,b=2k$$

So, $(2h)^2+(2k)^2=10^2\implies h^2+k^2=5^2$

So, the locus of $M(h,k)$ is $x^2+y^2=25$ which is evidently a circle centred at the origin with radius $5$ unit.

share|improve this answer
add comment

You should compute the analytical representation of the locus defined by your definition of M.

A = (x,0)
B = (0,sqrt(10^2-x^2))

M = (x/2, sqrt(10^2-x^2)/2)


Circle centered in 0 with radius 5: x^2+y^2 = 5^2

To prove that M is always on the circle you should substitute its x and y coordinates in the equation of the circle and observe that you obtain an identity.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.