Melody

If E is the middle point of the side CA of the triangle ABC and if R is the area of the triangle, prove that:

Cot AEB = (BC2 - BA2) / 4R  

Let

x=AB

y=CB

w=EB

z=AE=EC

angle AEB= theta

angle CEB= 180-theta

\(R=0.5zwsin\theta+0.5zwsin(180-\theta)\\ R=0.5zwsin\theta+0.5zwsin(\theta)\\ R=zwsin\theta\\ 4R=4zwsin\theta\\ \)

\(y^2=w^2+z^2-2wzcos(180-\theta)\\ y^2=w^2+z^2+2wzcos(\theta)\\ x^2=w^2+z^2-2wzcos(\theta)\\ subtracting\\ y^2-x^2=4wzcos(\theta)\\ BC^2-BA^2=4wzcos(\theta)\\ \)

\(\therefore\\ RHS =\frac{BC^2-BA^2}{4R}\\ RHS =\frac{4wzcos(\theta)}{4wzsin(\theta)}\\ RHS =cot(\theta)\\ RHS=LHS \qquad QED\)

Can we assume that  4R is meant to be in brackets?

Travisio, do not use this forum to get all your homework done for you - no matter what subject!

Do not ask any more of these questions in the remotely near future. 

Find a 2*2 matrix A such that       \(\binom{2 \text{ } 7}{5\text{ } 17}A= \binom{3 \text{ } -2}{4 \text{ } 1}\)

\(\binom{2 \text{ } 7}{5\text{ } 17}^{-1}=\frac{1}{2*17-7*5}\binom{17 \text{ } -7}{-5\text{ } 2}=\binom{-17 \text{ } 7}{5\text{ } -2}\)

\(A=\binom{-17 \text{ } 7}{5\text{ } -2}\binom{3 \text{ } -2}{4\text{ } 1}=\binom{-23 \text{ } 41}{7\text{ } -12}\)

Thanks guest.  Also thanks to EP and Hectictar for the earlier answers.

I will put time into digesting this when I get a long moment.

Thanks!   

You need to use more explanation, I have no idea what you are talking about.

Hi Tom,

You do not need to appologise for posting 2 questions, it is reasonably clear that you are using this forum as an an aid to your learning.

Off hand I do not see why there would need to be any quadratics with integer roots. Especially when we do not even know how many kids are in the class.

I could easily be wrong and I do not know how to show it one way or the other.

Since it could be a class of hundreds of students it is impossible to write out all the possibilities.

Huh??

ok, no worries, how did you do it?

Here is the graph