Finding Coordinates of A and B in an Equilateral Triangle

[ehild]

This also yields a simultaneous equation right?
Anyway, thank you so much! I learned a lot from this thread. Thanks everyone!

It yields the equations in bold in my previous post. See if you get the same ones, by applying the summation laws.

The equations are very easy to solve: There is only one unknown, a, in the first one.


ehild

 

[adjacent]

It yields the equations in bold in my previous post. See if you get the same ones, by applying the summation laws.

The equations are very easy to solve: There is only one unknown, a, in the first one.

I get the same thing too. Thanks for helping me.

 

[sankalpmittal]

It can be also done by using Coni's method. BTW, adjacent you're 15 ?

 

[thelema418]

I just want to point out that this problem has two possible solutions: I only saw one posted. You can see this by drawing lines to represent the possible points for A and B. In other words, $(a, 11)$ is somewhere on the line $y = 11$. The point $(b, 37)$ is somewhere on the line $y = 37$. Since the distance from O to A and from O to B is the same, we can draw a circle with radius $r$ and center at O. The intersections show the possible solutions. One is in QI and the other is in QII.

 

[sankalpmittal]

I just want to point out that this problem has two possible solutions: I only saw one posted. You can see this by drawing lines to represent the possible points for A and B. In other words, $(a, 11)$ is somewhere on the line $y = 11$. The point $(b, 37)$ is somewhere on the line $y = 37$. Since the distance from O to A and from O to B is the same, we can draw a circle with radius $r$ and center at O. The intersections show the possible solutions. One is in QI and the other is in QII.

Yes, by distance formula it's obvious that there are two solutions. To reduce the calculation part, I preferred Coni's method(in complex numbers), which also offers two solutions.z₁/z₂= e^iπ/3

In this case |z₁|=|z₂|...being an equilateral triangle.

 

[ehild]

I just want to point out that this problem has two possible solutions: I only saw one posted. You can see this by drawing lines to represent the possible points for A and B. In other words, $(a, 11)$ is somewhere on the line $y = 11$. The point $(b, 37)$ is somewhere on the line $y = 37$.

Yes, but the given picture suggest position in the first quadrant.

ehild

 

[ehild]

I just want to point out that this problem has two possible solutions: I only saw one posted. You can see this by drawing lines to represent the possible points for A and B. In other words, $(a, 11)$ is somewhere on the line $y = 11$. The point $(b, 37)$ is somewhere on the line $y = 37$.

Yes, but the given picture suggest position in the first quadrant.

ehild

 

[sankalpmittal]

Yes, but the given picture suggest position in the first quadrant.

ehild

So Coni's method will offer one solution as z₁/z₂=e^iπ/3.

Other will depend on quadrants due to the term e^iπ/3 (sign of angle). That won't matter as it is first quadrant related.