Analytic Geometry question I can't solve

Question

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?

Answer 1

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 $B=(11,6)$.

Answer 2

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