Q.Find the equation of the locus of a point which is equidistant from the points $A(2,3)$ and $B(6,-1)$.

The moving point $P$ satisfies $PA = PB$, so squaring both distances and simplifying will remove the square roots and leave a straight-line equation — this is always the perpendicular bisector of $AB$.

Let $P(x,y)$. Then $PA^2 = PB^2$:

Expanding: $x^2-4x+4+y^2-6y+9 = x^2-12x+36+y^2+2y+1$.

Cancelling $x^2,y^2$: $-4x-6y+13 = -12x+2y+37$, so $8x-8y-24=0$, i.e. $x-y-3=0$.

Let the moving point be $P(x,y)$. The condition given is that $P$ is equidistant from $A(2,3)$ and $B(6,-1)$, i.e. $PA = PB$, or equivalently $PA^2 = PB^2$ (squaring avoids the square roots in the distance formula).

Using the distance formula:

Setting them equal:

Expanding the left side: $x^2 - 4x + 4 + y^2 - 6y + 9$.

Expanding the right side: $x^2 - 12x + 36 + y^2 + 2y + 1$.

So: $-4x + 4 - 6y + 9 = -12x + 36 + 2y + 1$, i.e. $-4x - 6y + 13 = -12x + 2y + 37$.

Bringing all terms to one side: $-4x + 12x - 6y - 2y + 13 - 37 = 0 \Rightarrow 8x - 8y - 24 = 0$, which simplifies (dividing by $8$) to:

Independent check. Geometrically, this locus must be the perpendicular bisector of $AB$, so two things must both hold: (i) it passes through the midpoint of $AB$, and (ii) its slope is the negative reciprocal of the slope of $AB$. Midpoint of $A(2,3), B(6,-1)$ is $\left(\dfrac{2+6}{2}, \dfrac{3-1}{2}\right) = (4,1)$. Substituting into $x-y-3=0$: $4-1-3=0$ ✓. Slope of $AB = \dfrac{-1-3}{6-2} = \dfrac{-4}{4} = -1$, so the perpendicular bisector must have slope $1$; rewriting $x-y-3=0$ as $y=x-3$ confirms slope $1$ ✓. Both checks agree with the algebraic answer.

Alternative check: since the locus is known in advance to be the perpendicular bisector of $AB$, one could instead find the midpoint $(4,1)$ and the perpendicular slope $1$ directly, then write the line as $y-1=1(x-4)$, which also simplifies to $x-y-3=0$ — confirming the algebraic derivation without redoing the distance-formula expansion.

A common slip is expanding $(x-6)^2$ as $x^2-6x+36$ instead of $x^2-12x+36$ — always double the outer coefficient when squaring $(x-k)^2$. Another is forgetting to check the final line against the midpoint/perpendicularity property, which is the fastest way to catch a sign error in the expansion.