An isosceles triangle $ABC$ has 2 given vertices, $A(3,2)$ and $C (7,14$). The slope of AB is $\dfrac{1}{2}$. What are the coordinates of B?
I could figure out that line AB = $\dfrac{1}{2}x + \dfrac{1}{2} $
I found that the length of AC = is $\sqrt{160}$
But I haven't got a clue as to finding the coordinates of B.. can someone give me a hint?
Best Answer
Put $B(x,y)$. I solve your problem with assume the triangle $ABC$ isosceles at $A$. Because $AB =AC$, then $AB =\sqrt{160}$ or $$x^2+y^2-6x-4y-147 = 0.$$ The coordinates of the point $B$ are solutions of the system $$x^2+y^2-6x-4y-147 = 0, \quad y = \dfrac12x + \dfrac12.$$ We get $B(3 - 8\sqrt{2}, 2(1-2\sqrt{2})$ or $B(3 + 8\sqrt{2}, 2(1+2\sqrt{2})$.