Finding the locus in terms of a and b Find the locus of the point equidistant from $(a+b,b-a)$ and $(a-b,a+b)$ 
Okay so I don't know how do I find the locus of the point in terms of $ a$ & $b $.
 A: Hint: Write $a+b=x,a-b=y$. Now the point is equidistant. Let a point on the locus be $(\alpha,\beta)$ . Use the distance between two points formula and equate them.
A: ETA: This needs to be fixed.  ETA2: It has now been fixed.
Another way to look at this problem—one that in this case may be more illuminating—is geometrically.  Consider the points $C(a+b, b-a)$ and $D(a-b, a+b)$.  The midpoint of $\overline{CD}$ is $M(a, b)$, and the slope of $\overline{CD}$ is $\frac{2a}{-2b} = -\frac{a}{b}$.  Therefore, $\overline{CD}$'s perpendicular bisector—which is the locus of points equidistant from $C$ and $D$—must have a slope of $\frac{-1}{-a/b} = b/a$ and pass through the point $M$.  This is none other than the line
$$
y = \frac{b}{a}x
$$
A: If I understand right, you want the points that are the same distance from both given points. That is always a straight line, and for points $(c,d)$ and $(d,c)$ it's always $(x,x)$ for all real $x$. If you want it in terms of $a$ and $b$, then try to solve the equation $ax=a$ in terms of $a$ for $a\ne0$.
A: Let $(x,y)$ be a point on the locus. Then:
$$(x-(a+b))^2+(y-(b-a))^2=(x-(a-b))^2+(y-(a+b))^2$$
Expanding:
$$x^2-2x(a+b)+(a+b)^2+y^2-2y(b-a)+(b-a)^2=x^2-2x(a-b)+(a-b)^2+y^2-2y(a+b)+(a+b)^2$$
$$-2y(b-a)+2y(a+b)=2x(a+b)-2x(a-b)$$
$$4ya=4xb$$
$$y=\frac{b}{a}x$$
A: The other answers seems to be based on points (a+b, a-b) and (a-b, a+b), which would give the line y=x.
With the given points (a+b, b-a), (a-b, a+b), with same calculations as Ian Miller did, we get :
$-2y(b-a)+2y(a+b)=2x(a+b)-2x(a-b)$
which gives $ya=xb$, which is the same as the line $y=(b/a)x$
(line of slope $b/a$ — or direction $(a,b)$ — which passes through the point (0,0) )
