Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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
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 $B=(11,6)$.

share|cite|improve this answer
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 $B(3 + 8\sqrt{2}, 2(1+2\sqrt{2})$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.