Q.Find the equation of the locus of a point which is equidistant from the points $A(1, 2)$ and $B(3, 4)$.

Equidistance from two fixed points gives the perpendicular bisector; set $PA^2 = PB^2$ for a general point $P(x,y)$ and simplify.

Expanding and cancelling: $-2x-4y+5 = -6x-8y+25$, giving $4x + 4y - 20 = 0$, i.e. $x + y - 5 = 0$.

Let $P(x, y)$ be a general point on the locus with $PA = PB$, so $PA^2 = PB^2$:

Expand both sides:

Cancel $x^2$ and $y^2$:

Bring all terms to the left:

Divide by $4$: $x + y - 5 = 0$.

Check (independent). The midpoint of $AB$ is $(2, 3)$, and $2 + 3 - 5 = 0$, so it lies on the locus; the slope of $AB$ is $1$, so the perpendicular bisector has slope $-1$, matching $y = -x + 5$. ✓

Sign slips when expanding the squared brackets, and forgetting to divide the final equation by the common factor $4$ (leaving $4x + 4y - 20 = 0$ is not wrong, but $x + y - 5 = 0$ is the tidiest form).