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Could someone help me find the locations of the points A and B? Let me know if you need more information.

diagram

Thanks ;)

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  • 2
    $\begingroup$ This StackExchange is about the software system Mathematica (and how to use it), not for answering math questions. You should post this to the Math StackExchange. That said, if this is a homework assignment you really should try to figure it out yourself. $\endgroup$
    – Cassini
    Feb 22 at 21:21

2 Answers 2

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or = {0, 0};
Pl = 7/2 {-Sin[45 °], Cos[45 °]};
Pr = {0, 7/2};
Pm = Midpoint[{or, Pl}];
X = Midpoint[{Pl, Pr}];
B = FullSimplify[
   RegionIntersection[Line[{or, X}], Line[{Pr, Pm}]][[1, 1]]];
dBX = FullSimplify[EuclideanDistance[B, X]];
BX = FullSimplify[Normalize[X - B]];
A = FullSimplify[X + BX*dBX];
Graphics[{Circle[{0, 0}, 7/2], Line[{or, A}],
  Point[{Pl, Pr, Pm, A, B, X}], Opacity[.2], Triangle[{{0, 0}, Pr, Pl}]
  }]

A is at {-(7/(3 Sqrt[2])), 7/6 (2 + Sqrt[2])} and B is at {-(7/(6 Sqrt[2])), 7/12 (2 + Sqrt[2])}

enter image description here

This can be simplified further:

or = {0, 0};
Pl = 7/2 {-Sin[45 °], Cos[45 °]};
Pr = {0, 7/2};
Pm = Midpoint[{or, Pl}];
X = Midpoint[{Pl, Pr}];
B = FullSimplify@TriangleCenter[Triangle[{{0, 0}, Pr, Pl}], {"Centroid"}];
A = 2 B;
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Name the points as in the graphics below. The we can write:

Clear["Global`*"]
ab = ac = 3.5;
a = {0, 0};
b = {0, 3.5};
c = 3.5  {-Sin[45  Degree], Cos[45  Degree]};
d = c/2;
e = (b + c)/2;

Then B lies on ae, a fraction "l1" from a. And B lies also on db a fraction l2 from b. This gives rise to the equations:

eq = {l1  (e - a) == l2  (d - b) + b};

With this:

B = l1  (e - a) /. Solve[eq, {l1, l2}][[1]]
A = (1 + 1 - l1) e /. Solve[eq, {l1, l2}][[1]]

{-0.824958, 1.99162}
{-1.64992, 3.98325}

And the graphics:

enter image description here

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