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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?

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That may very well be true. What exactly doesn't make sense? –  JohnPhteven Oct 6 '12 at 15:14
    
Oh, changed it, thanks for noticing! –  JohnPhteven Oct 6 '12 at 15:20

2 Answers 2

up vote 0 down vote accepted

The first thing to do is to make a reasonably accurate sketch.

It looks as if there are three triangles, two of which were identified by min_thao2011. We might also have $BC=BA$. If we let the coordinates of $B$ be $(x,y)$, this yields the equation $$(x-7)^2+(y-14)^2=(x-3)^2+(y-2)^2.$$ There is useful cancellation. Combine with the known equation for the line $AB$. We get $B=(11,6)$.

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I want to do it the way you say in the last sentence. I get as a final equation $\dfrac{1}{2}x + \dfrac{1}{2} = 16\dfrac{1}{3} - \dfrac{1}{3}x $ But this yields totally different results. Would you mind telling me what I do wrong? –  JohnPhteven Oct 6 '12 at 16:59
    
I had the wrong line in the comment at the end. So will just stick with the one way. –  André Nicolas Oct 6 '12 at 17:16
    
But if you take the point on line AB equidistant from A and B you get (11,6) as your solution, which is logical because AB=BC! –  JohnPhteven Oct 6 '12 at 17:50
    
@ZafarS: yes, good observation. I think it is the other answer that should be accepted, it produced two of the three possibilities! –  André Nicolas Oct 6 '12 at 18:10

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 $B(3 + 8\sqrt{2}, 2(1+2\sqrt{2})$.

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