Q.If A(1,3) and B(2,1) are points, find the equation of the locus of point P such that PA = PB.

Let $P(x,y)$ be any point on the required locus. Since $PA = PB$, we also have $PA^2 = PB^2$:

Expanding both sides: $x^2-2x+1+y^2-6y+9 = x^2-4x+4+y^2-2y+1$.

Cancelling $x^2,y^2$ and simplifying: $-2x-6y+10=-4x-2y+5$.

Bringing all terms to one side: $2x-4y+5=0$, i.e. $x-2y+2.5=0$; dividing to standard integer form gives $2x-4y+5=0$, which simplifies no further, so we keep it as $2x-4y+5=0$, equivalently $x-2y+2.5=0$. Re-checking the arithmetic: $-2x-6y+10=-4x-2y+5 \Rightarrow -2x-6y+10+4x+2y-5=0 \Rightarrow 2x-4y+5=0$.

Set the distance condition PA = PB as PA² = PB², expand both squared-distance expressions, and simplify.

• Forgetting to square the distances before expanding, which leaves an awkward square-root equation.
• Sign errors when expanding $(x-1)^2$ vs $(x-2)^2$.
• Not cancelling the $x^2$ and $y^2$ terms, which always cancel in a PA=PB locus and should reduce it to a straight line.