Locus

What a locus is. A locus is the path traced by a point moving under a rule, or equivalently the set of all points in the plane satisfying a given condition — no more, no fewer. The defining property runs both ways: every point that satisfies the equation lies on the locus, and every point on the locus satisfies the equation. Getting the algebra to match the geometry exactly, in both directions, is the entire point of the topic.

The four-step method. Every locus problem follows the same template.

• Let $P(x,y)$ be a general point assumed to lie on the locus.
• Translate the stated condition into an equation in $x$ and $y$, using the distance formula, slope, or section formula as needed.
• Simplify the equation as far as it will go.
• Sanity-check with a point that must lie on the locus by the geometry of the situation (a mid-point, for instance).

The distance between two points $(x_1,y_1)$ and $(x_2,y_2)$ is the workhorse tool behind almost every locus condition:

where: $d$ is the distance between the two points and $(x_1,y_1)$, $(x_2,y_2)$ are their coordinates. Squaring both sides of a distance equation is always legitimate here, since a distance is never negative — squaring loses no solutions and introduces none, so it is the standard way to clear the square root before expanding.

Three conditions you already know the answer to. Certain wordings are common enough to recognise on sight, though a locus problem still expects the full algebraic derivation:

• Equidistant from two fixed points $\to$ the perpendicular bisector of the segment joining them (a straight line).
• A fixed distance $r$ from a fixed centre $\to$ a circle of radius $r$ (the standard circle equation is exactly this locus, derived the same way).
• Equidistant from two fixed lines $\to$ one of the angle bisectors of those lines.

Recognising the pattern is a useful check on your final simplified equation, but it is never a substitute for doing the algebra — problems routinely disguise a simple condition (like "twice as far") that produces something other than a line.

Beyond distance. Not every locus condition is stated purely in terms of distance. Some conditions are naturally phrased using slope — for instance, a point moving so that the line joining it to a fixed point always makes a fixed angle with the x-axis — while others involve the section formula, when a moving point is defined as dividing a segment between two other moving or fixed points in a given ratio. The four-step method still applies unchanged; only the specific tool used in Step 2 to convert words into algebra differs. This is why locus sits at the end of the Straight Lines unit rather than as a standalone topic — it deliberately calls back on everything else the chapter has already built.

Find the equation of the locus of a point $P$ equidistant from $A(3,4)$ and $B(-1,2)$, then verify it.

Step 1. Set up the condition. Let $P(x,y)$ satisfy $PA = PB$, so $PA^2=PB^2$:

Step 2. Expand both sides:

Step 3. Cancel $x^2$ and $y^2$, then collect terms onto one side:

Step 4. Divide through by $-4$ to reach the simplest form: $2x+y-5=0$.

Step 5. Verify using the midpoint of $AB$, which must lie on the locus since it is equidistant from both endpoints by construction: midpoint $=\left(\frac{3-1}{2},\frac{4+2}{2}\right)=(1,3)$, and $2(1)+3-5=0$ ✓. The locus is the line $2x+y-5=0$, the perpendicular bisector of $AB$.

Competitive stretch

A point $P$ moves so its distance from $(2,0)$ is always twice its distance from $(0,1)$: writing $PA^2=4\,PB^2$ and expanding gives $x^2-4x+4+y^2 = 4(x^2+y^2-2y+1)$, which simplifies to $3x^2+3y^2+4x-8y=0$, or $x^2+y^2+\frac{4}{3}x-\frac{8}{3}y=0$ after dividing by 3 — a circle, since the $x^2$ and $y^2$ terms no longer cancel once the ratio is unequal.

• Squaring only one side, or squaring before isolating the distances. Both sides of $PA=PB$ (or $PA=k\,PB$) must be squared together to clear all square roots cleanly in one step.
• Sign errors expanding $(x-a)^2$ or $(y-b)^2$. A dropped negative sign inside one of several expansions in the same problem is the single most common cause of a wrong final line or circle.
• Stopping before full simplification. An unsimplified but correct locus equation can look "wrong" next to a textbook answer even though the two are algebraically identical — always divide out common factors.
• Swapping which point carries the multiplier in an unequal-ratio condition. $PA=2PB$ and $PB=2PA$ describe two different circles; misreading which distance is doubled changes the entire answer.

Ask the tutor: If a worded locus condition is confusing you, describe it to the AI tutor and ask for help writing the very first equation — that translation step is where most of the difficulty lives.