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Mathematics · Ch 5 — Straight Line

Equation of Locus

5.1.1

Equation of Locus

5.1.1 Equation of Locus

Suppose every point on a locus has coordinates satisfying some algebraic equation in xx and yy, and — just as importantly — no point off the locus satisfies that equation. Then that equation is called the equation of the locus.

Worked Example 1. Every point on the X-axis has yy-coordinate 00, and this is true only for points on the X-axis (no off-axis point has y=0y=0). So the equation of the X-axis is simply

y=0.y = 0.

Worked Example 2. Let L={P∣OP=4}L = \{P \mid OP = 4\}; find its equation. Take a general point P(x,y)P(x,y) on LL. Since OP=4OP = 4, we also have OP2=16OP^2 = 16, so by the distance formula

(x−0)2+(y−0)2=16  ⟹  x2+y2=16.(x-0)^2 + (y-0)^2 = 16 \implies x^2+y^2 = 16.

This is the equation of the locus LL, and geometrically the locus is a circle of radius 4 centred at the origin.

Worked Example 3. Find the equation of the locus of points equidistant from A(−3,0)A(-3,0) and B(3,0)B(3,0), and identify it. Let P(x,y)P(x,y) be any point on the required locus. Since PP is equidistant from AA and BB, PA=PBPA = PB, hence PA2=PB2PA^2 = PB^2:

(x+3)2+y2=(x−3)2+y2.(x+3)^2 + y^2 = (x-3)^2 + y^2. …

Misc 5.1.1-Ex1Ex. 1 — equation of the X-axis

Worked out. Reasons that the y-coordinate of every point on the X-axis is 0, and only points on the X-axis have y = 0, so y = 0 is the equation of the X-axis. Working through this worked example after reading the theory above helps consolidate the method before attempting the exercise questions that follow it in th …

Misc 5.1.1-Ex2Ex. 2 — locus of points 4 units from the origin

Worked out. Finds the equation of the set of points at a fixed distance 4 from the origin by letting P(x,y) be a general point, using OP = 4, and squaring to reach x² + y² = 16; identifies the locus as a circle. …

Figure 5.1.1-Fig5.2Fig. 5.2

What this figure shows. Diagram showing the circular locus of Example 2 — the origin O and a general point P at distance 4 from it. This figure gives the reader a concrete visual reference for the geometric configuration described in the surrounding text, tying the abstract statement to a p …

Misc 5.1.1-Ex3Ex. 3 — locus equidistant from A(−3,0) and B(3,0)

Worked out. Sets PA = PB for a general point P(x,y), squares both sides, expands, and cancels common terms to find the equation of the locus; identifies the resulting locus as the Y-axis. …