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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 $.

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    $\begingroup$ Those are fwo points. Let's name them: point $A(a+b,b−a)$ and point $B(a−b,a+b)$. Points equidistant from $A$ and $B$ form a line – a bisector of the $\overline{AB}$ segment. $\endgroup$ – CiaPan Oct 21 '15 at 6:14
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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

$$ y = \frac{b}{a}x $$

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  • $\begingroup$ $C$ is actually $(p,-q)$ as the question is written $\endgroup$ – David Z Oct 21 '15 at 9:41
  • $\begingroup$ Good grief—I'll have to fix this tomorrow. $\endgroup$ – Brian Tung Oct 21 '15 at 9:47
  • $\begingroup$ @DavidZ: Fixed. Thanks for the sharp eye and seeing what the rest of us (outside of Anceps) missed. $\endgroup$ – Brian Tung Oct 21 '15 at 17:25
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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.

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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 $a\ne0$.

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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) )

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  • $\begingroup$ +1 for catching the asymmetry the rest of us (except for David Z) missed. $\endgroup$ – Brian Tung Oct 21 '15 at 17:25
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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: $$x^2-2x(a+b)+(a+b)^2+y^2-2y(b-a)+(b-a)^2=x^2-2x(a-b)+(a-b)^2+y^2-2y(a+b)+(a+b)^2$$ $$-2y(b-a)+2y(a+b)=2x(a+b)-2x(a-b)$$ $$4ya=4xb$$ $$y=\frac{b}{a}x$$

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  • $\begingroup$ Corrected based on comments from Anceps and others. $\endgroup$ – Ian Miller Oct 22 '15 at 0:38

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