The points A(5,1) and B(-3,6) represent one of the equal sides of an isosceles triangle. Determine one of the possible points that would represent the third vertex of the triangle. Provide calculations to support your answer.

Thanks

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- Jul 10th 2008, 09:29 PMeuclid2Triangle vertex
The points A(5,1) and B(-3,6) represent one of the equal sides of an isosceles triangle. Determine one of the possible points that would represent the third vertex of the triangle. Provide calculations to support your answer.

Thanks - Jul 10th 2008, 09:44 PMearboth
The third vertex of the triangle is located on a circle around A with radius $\displaystyle r = \sqrt{89}$ , that means the circle passes through B.

OR

the third vertex of the triangle is located on a circle around B with radius $\displaystyle r = \sqrt{89}$ , that means the circle passes through A.

EDIT: I've repaired a silly mistake. Thank you, kalagota! - Jul 10th 2008, 09:46 PMkalagota
first, compute the distance between A and B.

second, the set of all points that satisfies you condition is the set of (x,y) such that the distance between A (or B) and (x,y) is equal to the distance between A and B.

so for example, $\displaystyle d^2 = (x_B-x_A)^2 + (y_B-y_A)^2 = 89$

we consider: from point A, we will find a point (x,y) such that the distance of A and (x,y) is $\displaystyle d$.

so $\displaystyle d^2 = 89 = (x-5)^2 + (y-1)^2$ (this equation represents a circle) the point is, you can find any (x,y) that satisfies that equation..

NOTE: if you find an (x,y), be sure that A,B and (x,y) are not collinear.

you can also do this with respect to point B. so, there will be a new equation.. with same d. - Jul 10th 2008, 09:48 PMkalagota
- Jul 10th 2008, 09:52 PMearboth
In addition to my previous post I've attached a sketch of the situation.

I've choosen a point on the circle $\displaystyle c_A$ (approximately (-8, -2) check if this point lies**exactly**on this circle!)

and

a point on the circle $\displaystyle c_B$ (approximately (0,-7) check if this point lies**exactly**on this circle!) - Jul 10th 2008, 10:03 PMkalagota
- Jul 10th 2008, 10:15 PMeuclid2
are you suggesting that either one of those points are appropriate for the third vertex, if so how did you come up with those? and how do i verify they are correct?

- Jul 10th 2008, 10:21 PMearboth
You only have to use the equations kalagota has posted in this thread:

Take (-8, -2) and plug in the coordinates into the appropriate equation:

$\displaystyle (x+3)^2+(y-6)^2=89$ will yield:

$\displaystyle (-8+3)^2+(-2-6)^2=89$ Check! So (-8,-2) is indeed a vertex of the isosceles triangle. - Jul 10th 2008, 10:24 PMeuclid2
Ok thanks.

I'm just not sure where you get the (-8,-2) from that's all? - Jul 11th 2008, 01:18 AMearboth
If you have a look at the drawing I posted in one of my previous post you'll probably find some points with integer coordinates which lie on one of the circles. These points are my first guess.

Since a drawing isn't accurate enough I advised you to check, if the point lies**exactly**on the circle by using the equations of the circle. That's all, nothing mysterious about it.