Coordinate Geometry(Right Triangle)

In summary: Hi Ritwik! Thanks for the PM. :smile: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. :wink:
  • #1
ritwik06
580
0

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
 
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  • #2
Try using www.imageshack.us or create an account on photobucket. They support direct links.
 
  • #3
Please help me with the question!
 
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  • #4
Hi Ritwik! Thanks for the PM. :smile:

(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. :smile:
 
  • #5
tiny-tim said:
Hi Ritwik! Thanks for the PM. :smile:

(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. :smile:

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?
 
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  • #6
Hi Ritwik! :smile:

I followed you down to …
ritwik06 said:
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 = … ? :wink:
 
  • #7
tiny-tim said:
Hi Ritwik! :smile:

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 = … ? :wink:
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
ritwik06 said:
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. :wink:
 

Related to Coordinate Geometry(Right Triangle)

1. What is the Pythagorean Theorem?

The Pythagorean Theorem states that in a right triangle, the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the other two sides.

2. How do you find the length of a side in a right triangle?

To find the length of a side in a right triangle, you can use the Pythagorean Theorem or the trigonometric ratios (sine, cosine, and tangent) depending on the given information.

3. What is the relationship between the sides and angles in a right triangle?

In a right triangle, the sides are related by the Pythagorean Theorem and the angles are related by the trigonometric ratios. The sine of an angle is equal to the ratio of the opposite side to the hypotenuse, the cosine is equal to the adjacent side to the hypotenuse, and the tangent is equal to the opposite side to the adjacent side.

4. How do you determine if a triangle is a right triangle?

If the triangle has one angle that measures 90 degrees, then it is a right triangle. This angle is also known as the right angle.

5. Can you find the area of a right triangle using coordinate geometry?

Yes, you can find the area of a right triangle using the formula A = 1/2 * base * height. The base and height can be determined by using the coordinates of the vertices of the triangle and applying the distance formula.

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