Coordinate Geometry(Right Triangle)

1. Feb 12, 2009

ritwik06

1. The problem statement, all variables and given/known data
If y=x+2, y=2x+3 are medians of right angled triangle ABC (angle A=90) through B and C respectively such that |BC|=81 units. Find area of triangle.

3. The attempt at a solution
I have been trying this since a long time.

I am posting my work in the form of 2 images. I hope my writing is legible. I think I have generated enough equations for the unknowns. But to solve them is difficult. Please help me.

I was also told that there exists a shorter method than this. I shall be glad if you could guide me to a shorter approach.

Picture 1:
http://img18.imageshack.us/img18/4679/87911896fu2.jpg [Broken]

Picture 2:

http://img25.imageshack.us/img25/694/76694925pt3.jpg [Broken]

regards,
Ritwik

Last edited by a moderator: May 4, 2017
2. Feb 12, 2009

Mezzlegasm

Try using www.imageshack.us or create an account on photobucket. They support direct links.

3. Feb 12, 2009

ritwik06

Please help me with the question!!

Last edited: Feb 12, 2009
4. Feb 13, 2009

tiny-tim

Hi Ritwik! Thanks for the PM.

(I find it very difficult to read the pictures of your handwriting: it would be easier if you would type your work directly into your post)

Hint: the only importance of the x and y coordinates is that they tell you the angle between the medians.

Work out that angle, and then forget all about the coordinates, just draw a general right-angled triangle ABC, and work out what the area is, given the length BC and the angle between the medians.

5. Feb 13, 2009

ritwik06

Hi Tim,
Its me who should thank you.
I am very grateful that you replied.
Your advice was very-very useful. I have been able to solve the question. Now I am working towards a more compact and smarter solution. I hope you will help :P

Here is how I got the result.
Let O be the angle between the medians.
Then tan O=(1/3)

Area of triangle EMF=(1/12) *Area of the whole triangle(A)

0.5*(1/3)*M1*(1/3)*M2=(1/12)*A

Now in the same triangle EMF;
I apply the cosine rule
cos(pi-O)=$$\frac{((1/3)*M1) ^2 +((1/3)*M2)^2)-(a/2)^2}{\frac{2M1M2}{9}}$$

After putting values of cos(pi-O), M1*M2. I still have to find M1^2+ M2^2
Applying cosine rule;
CF^2(M1^2)=BC^2+BF^2-2*BC* BF* cos B

Again putting values of 2*BC* BF* cos B from the original cosine rule equation. for triangle ABC

Thn I plug values of M1^2+ M2^2 in the original equation. an I get the area=729 sq units.
Now the question arises. Is there a smaller time saving method???

Last edited: Feb 13, 2009
6. Feb 13, 2009

tiny-tim

Hi Ritwik!

I followed you down to …
… and then I got a bit lost.

But you could speed things up by using Pythagoras: M12 + M22 = … ?

7. Feb 14, 2009

ritwik06

Then I again used the cosine rule there:
to replace 2*BC*BF*cos B from this expression,
2*ac*cos B=a^2+c^2-b^2

But as you said pythagoras was a better option. Thanks a lot.
I was wondering whether there could be a method without involving the areas of the smaller triangles as I did???

By the way thanks a lot for your help :D
and Happy Valentines Day!!!

8. Feb 15, 2009

tiny-tim

You could use vectors instead …
put the origin at O = A, so that OB = b, OC = c.

Then the medians are (b/2 - c) and (c/2 - b), and b.c = 0

so cos = (b/2 - c).(c/2 - b)/√((b/2 - c)2(c/2 - b)2)

which you can turn into an equation in bc by using b2 + c2 = 812, and by completing the square.

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