Locus of a point

What a locus is

A locus is the path of a point that keeps to one rule. A point that stays 5 cm from a pin traces a circle. A point that stays equally far from two pins traces a straight line. Straight-line problems in JEE and in A-Level Further Maths often give a rule like this and ask for the line.

Three steps find the equation of a locus.

1. Name the point. Call the moving point $P(x, y)$. The fixed points and lines keep their own coordinates.
2. Write the rule. Turn the words into an equation in $x$ and $y$. A distance contains a square root, so square both sides to remove it.
3. Simplify and look. Collect the terms, say what shape you have, and note any point that must be left out.

Equal distances from two points

The rule $PA = PB$ is the same as $PA^2 = PB^2$, and squaring removes the square roots.

Example 1. Find the locus of $P$ when $PA = PB$, with $A(-3, 2)$ and $B(5, 8)$.

Expand both sides. The $x^2$ and $y^2$ terms cancel, and a line is left:

Check it. The middle point of $AB$ is $M(1, 5)$, and $4 + 15 = 19$. The slope of $AB$ is $\dfrac{3}{4}$, and the slope of the locus is $-\dfrac{4}{3}$. Their product is $-1$. So the locus is the perpendicular bisector of $AB$.

For any $A(x_1, y_1)$ and $B(x_2, y_2)$ the same steps give

Distances from two lines

When the rule is about distances from lines, use the formula $\dfrac{|ax + by + c|}{\sqrt{a^2 + b^2}}$. The modulus signs matter, because removing them gives two cases.

Example 2. The distance of $P$ from the line $3x + 4y = 10$ is twice its distance from the $y$-axis. Find the locus.

The distance from the $y$-axis is $|x|$. So

The plus sign gives $7x - 4y + 10 = 0$. The minus sign gives $13x + 4y - 10 = 0$. The locus is a pair of lines, and both pass through $(0, 2.5)$, where the given line meets the $y$-axis.

A fixed ratio of distances from two lines always gives a pair of lines through the point where they meet. A fixed sum of distances gives a closed shape.

Example 3. The sum of the distances of $P$ from the two axes is 5. Then $|x| + |y| = 5$. In the first quadrant this is $x + y = 5$, and the other three quadrants give the mirror images. The locus is a square with corners $(\pm 5, 0)$ and $(0, \pm 5)$. Its diagonals are 10 long, so its area is $\dfrac{10 \times 10}{2} = 50$.

Conditions on slopes

Slopes help when the rule is about angles at $P$. The slope of $PA$ is $\dfrac{y - y_1}{x - x_1}$, which does not exist at $A$ itself. So the fixed points are left out of the locus.

Example 4 (a right angle). $A(-2, 1)$ and $B(6, 7)$. The point $P$ moves so that $PA \perp PB$. The product of the two slopes is $-1$:

Expanding gives $x^2 + y^2 - 4x - 8y - 5 = 0$, which is $(x - 2)^2 + (y - 4)^2 = 25$. The locus is the circle with $AB$ as its diameter, without the points $A$ and $B$. This agrees with the fact that the angle in a semicircle is a right angle.

Example 5 (slopes that add to zero). $A(1, 0)$ and $B(5, 0)$. The slopes of $PA$ and $PB$ add to zero:

So $y = 0$ or $x = 3$. The locus is the vertical line $x = 3$ together with the line $AB$ itself, leaving out $A$ and $B$. A fixed sum of slopes that is not zero gives a hyperbola, which belongs to a later chapter.

A moving line and its midpoint

Sometimes the moving point is made from a line that changes. Name the changing numbers, write the point in terms of them, and then remove them.

Example 6. A line cuts the axes at $A(a, 0)$ and $B(0, b)$ with $2a + b = 20$. Find the locus of the middle point $M$ of $AB$.

Here $M = (x, y) = \left(\dfrac{a}{2}, \dfrac{b}{2}\right)$, so $a = 2x$ and $b = 2y$. Put these into the rule: $4x + 2y = 20$, which is $2x + y = 10$. The locus is a line with intercepts 5 and 10, the rule shrunk to half its size.

If the rule had been $a^2 + b^2 = 100$, a rod of fixed length sliding between the axes, the same steps would give $x^2 + y^2 = 25$, a circle.

Practise with new numbers

Try it in Python

The program checks the line $PA = PB$ for random pairs of points, and then the midpoint of a sliding line.

Common mistakes

• Squaring one side only. $PA = PB$ becomes $PA^2 = PB^2$. Both sides must be squared.
• Dropping the modulus. A distance from a line has $|\ \ |$ around it, so a ratio rule gives two lines, not one.
• Using the end point instead of the midpoint. Write $a$ and $b$ in terms of the $x$ and $y$ of the midpoint.
• Forgetting the points that must go. A slope rule fails at the fixed points themselves.

Remember these

• Call the moving point $P(x, y)$, write the rule, square to clear roots, then simplify.
• $PA = PB$ gives the perpendicular bisector of $AB$.
• A fixed ratio of distances from two lines gives a pair of lines through their meeting point.
• $PA \perp PB$ gives the circle on $AB$ as its diameter.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. Find the locus of points equally far from $(0, 4)$ and $(6, 0)$.

2. The sum of the distances of $P$ from the two axes is 7. What shape is the locus, and what is its area?

3. A line cuts the axes at $(p, 0)$ and $(0, q)$ with $3p + q = 18$. Find the locus of the midpoint.

4. $A$ is $(0, 0)$ and $B$ is $(12, 16)$. The point $P$ moves so that $PA \perp PB$. Describe the locus.