Coordinate Geometry(Right Triangle)

[ritwik06]

Homework Statement

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.

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 Picture 2:

http://img25.imageshack.us/img25/694/76694925pt3.jpg regards,
Ritwik

 

[Mezzlegasm]

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

 

[ritwik06]

Please help me with the question!

 

[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.

 

[ritwik06]

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.

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)=[tex]\frac{((1/3)*M1) ^2 +((1/3)*M2)^2)-(a/2)^2}{\frac{2M1M2}{9}}[/tex]

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?

 

[tiny-tim]

Hi Ritwik!

I followed you down to …

CF^2(M1^2)=BC^2+BF^2-2*BC* BF* cos B

… and then I got a bit lost.

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

 

[ritwik06]

Hi Ritwik!

I followed you down to …


… and then I got a bit lost.

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

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!

 

[tiny-tim]

I was wondering whether there could be a method without involving the areas of the smaller triangles as I did?

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.