Q.A point $P$ moves in the plane so that it is always equidistant from the two fixed points $A(2, 3)$ and $B(4, 1)$. Find the equation of the locus of $P$.

The condition "equidistant from two fixed points" describes the perpendicular bisector of the segment $AB$; take a general point $P(x,y)$ and set $PA = PB$, then square and simplify.

$(x-2)^2+(y-3)^2 = (x-4)^2+(y-1)^2$. Expanding and cancelling $x^2, y^2$: $-4x+4-6y+9 = -8x+16-2y+1$, which gives $4x - 4y - 4 = 0$, i.e. $x - y - 1 = 0$.

Let $P(x, y)$ be any point on the locus. The condition is that its distance from $A(2,3)$ equals its distance from $B(4,1)$, so $PA = PB$, and squaring both sides (which removes the square roots of the distance formula) gives $PA^2 = PB^2$:

Expand each side:

The terms $x^2$ and $y^2$ appear on both sides and cancel:

Bring all terms to the left:

Dividing by $4$: $x - y - 1 = 0$.

Check (independent). The locus should be the perpendicular bisector of $AB$. The midpoint of $AB$ is $\left(\tfrac{2+4}{2}, \tfrac{3+1}{2}\right) = (3, 2)$, and $3 - 2 - 1 = 0$, so the midpoint lies on our line. The slope of $AB$ is $\tfrac{1-3}{4-2} = -1$, so the perpendicular bisector has slope $+1$; our line $x - y - 1 = 0$ rearranges to $y = x - 1$, slope $1$. Both facts match, confirming the result.

Alternatively, use the known geometric fact directly: the locus of points equidistant from $A$ and $B$ is the perpendicular bisector of $AB$. Find the midpoint $(3,2)$ and the perpendicular slope $+1$, then apply the point-slope form $y - 2 = 1(x-3)$, giving $x - y - 1 = 0$ immediately.

Forgetting to square both sides leaves square roots that are awkward to simplify; and a sign slip when expanding $(y-3)^2 = y^2 - 6y + 9$ (writing $+6y$) changes the whole answer. Always expand each bracket fully before cancelling.