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How to prove that locus of point(P) whose distance from two fixed points(A,B) lie on right bisector. My textbook writes as " obviously all positions of moving point P lies on right bisector of AB". how do i see this?

thanks

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    Missing some info here. "whose distances from two fixed points are equal" would make sense. As in the note above, APB is always isosceles and therefor if you let M = midpoint AB then AM forms two congruent triangles. angle PMA = angle PMB, so AM must be perpendicular to AB. Draw this out and write a formal proof to convince yourself -- a valuable exercise. – 2017-02-06

2 Answers 2

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Let the two points be A(x1,y1) and B(x2,y2) respectively and point P be (x,y).

Slope of AB = m1= $ (y1-y2)/(x1-x2) $

Since the point P is equidistant from A and B,

$PA=PB$

or $PA^2=PB^2$

Therefore, $(x-x1)^2+(y-y1)^2=(x-x2)^2+(y-y2)^2$

On solving this, you get,

$h(x1-x2)+k(y1-y2)+c=0$

Here, slope of this line, $m2= -(x1-x2)/(y1-y2)$

Clearly, $m1.m2=-1$

Therefore, this line is perpendicular to the line joining A and B and any point on this line is equidistant from the point A and B.

Therefore, the locus of a point which is equidistant from two points is the perpendicular bisector of those two points.

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I am providing with two proofs: the first one basically uses purely Euclidean Geometry while the other is done upon the Cartesian plane.

Euclidean proof : Let the two given points be $A$ and $B$. $O$ be a point on the locus. Draw a perpendicular onto $AB$ at $P$ from $O$. $OA=OB \implies PO^2+PA^2=OA^2=OB^2=PO^2+PB^2 \implies PA=PB$. The point $O$ was basically chosen without any condition except the fact that it belongs to the locus. And it faithfully returns only one condition that is $PA=PB$ for any point $P$ such that $OP \perp AB$.

Now, take points $A,B$. Draw a right bisector. Now take any point $X$ that does not belong to the bisector. Draw $XQ \perp AB$. Argue like that and prove $XA \neq XB$ and this again means any point not on the right bisector isn't faithful, i.e., it won't lie on the pre-defined locus.

Coordinate proof : Assign $A(a_x,a_y), B(b_x,b_y)$. Let $O$ belong to the locus of equidistant points. Use the distance formula (basically Pythag) to get $(y-b_y)^2+(x-b_x)^2=(y-a_y)^2+(x-a_x)^2 \implies \frac{y-\frac{1}{2}(a_y+b_y)}{x-\frac{1}{2}(a_x+b_x)} \cdot \frac{b_y-a_y}{b_x-a_x}=-1$ which again, faithfully implies that $O$ belongs to the right bisector as the midpoint of $AB$ is basically $\bigg(\frac{1}{2}(a_x+b_x), \frac{1}{2}(a_y+b_y)\bigg)$.

Done!!