Problem regarding lines (Analytic Geometry)

Question

Here is the problem:

The base of a triangle has a fixed position and its length is constant and measures a. The difference of the squares of the other two sides is constant and measures $b^2$. Prove that the geometric space is a line.

Honestly, I cannot see where to start. The topic is "The Line", Analytic Geometry. I would appreciate any help as for how to interpret the problem.

Solution Since $a$ is a fixed line, we may give its start and end points names of $A(x_1,y_1)$ and $C(x_2,y_2)$. The third point, however, is not fixed, so it will be named $B(x,y)$. Now we go to the definition.

$$ d^2(A,B) = (x-x_1)^2 + (y-y_1)^2 \\ d^2(C,B) = (x-x_2)^2 + (y-y_2)^2 \\ b^2 = (x-x_1)^2 + (y-y_1)^2 - (x-x_2)^2 - (y-y_2)^2 \\ b^2 = x^2 + x_1^2 + 2xx_1 + y^2 + y_1^2 + 2yy_1 - x^2 - x_2^2 + 2xx_2 - y^2 - y_2^2 + 2yy_2 \\ b^2 = x_1^2 + 2xx_1 + y_1^2 + 2yy_1 - x_2^2 + 2xx_2 - y_2^2 + 2yy_2 $$ Now we have to put the whole thing in terms of $y$. $$ y(2y_1+2y_2) = x(-2x_1-2x_2) + x_2^2 + y_2^2 - x_2^2 - y_2^2 \\ y = x\frac{(-2x_1-2x_2)}{(2y_1+2y_2) } + \frac{x_2^2 + y_2^2 - x_2^2 - y_2^2}{(2y_1+2y_2)} $$

Well, there is the line. A simpler case would be with $P(0,0)$ and $Q(a,0)$. $$ b^2 = (x^2+y^2)-((x-a)^2 + y^2) \\ b^2 = 2ax \\ x = \frac{b^2}{2a} $$

Answer 1

Hint: Suppose two points on the base are $(x_0,y_0)$ and $(x_1,y_1)$ and the other point is $(x,y)$. The difference of squares of two other sides are difference of $(x-x_0)^2+(y-y_0)^2$ and $(x-x_1)^2+(y-y_1)^2$ which is $$ \left|(2x_1-2x_0)x+(2y_1-2y_0)y+2x_0x_1+2y_0y_1+a^2\right|=b^2. $$

Answer 2

If you choose the axes $(A, \vec{i}, \vec{j})$ with $\vec{i} = \dfrac{\overrightarrow{AB}}{AB}$ you will have the simple coordinates $A(0, 0)$ and $B(a, 0)$.
Let $B(x, y)$ then : $$b^2 = AB^2 - BC^2 = x^2 + y^2 - \left((x - a)^2 + y^2\right) = a (2 x - a)$$ We deduce that : $$x = \dfrac{1}{2} \left(\dfrac{b^2}{a} + a\right) = \dfrac{b^2 + a^2}{2 a}$$
Wich is the equation of a perpendicular line to the line $(AB)$.