LOCI A locus (pl. loci) is a set of points (ordered pairs) in a plane that satisfies some given condition. In other words, it is the path traced out by a point moving according to some pre-determined rule. A locus is usually described by an equation. Examples 1. Find the locus of the point that moves such that it is equidistant from the points A (8, -4) and B (-6, 2). Let the point be P (x, y) So, ∴ ∴ ∴ ∴ ∴ 2. PA = + (𝑦 − = √(𝑥 − (𝑥 − 8)2 + (𝑦 + 4)2 = x 2 – 16x + 64 + y 2 + 8y + 16 = 28x – 12y – 40 = 7x – 3y – 10 = 8)2 −4)2 PB √(𝑥 − −6)2 + (𝑦 − 2)2 (𝑥 + 6)2 + (𝑦 − 2)2 x 2 + 12x + 36 + y 2 – 4y + 4 0 0 Show that the locus above is the perpendicular bisector of AB. ∴ ∴ 7x – 3y – 10 3y y So, Now, ∴ = = = 0 7x – 10 7 10 𝑥− 3 3 m1 = 7 3 m AB = m 1 × m AB 𝑦2 − 𝑦1 𝑥2 − 𝑥1 = 2−−4 −6−8 = − 3 7 = 7 3 × − = -1 3 7 The locus is perpendicular to AB. The midpoint of AB = (1, -1) Substituting x = 1 into the equation of the locus, y = ∴ 7 10 𝑥− 3 3 7 10 − 3 3 = = -1 The midpoint is on the locus and hence the locus is the perpendicular bisector of AB. 3. Determine the locus of the point P (x, y) that is always 4 units from the y-axis. y A diagram easily illustrates the possible positions of the locus. It must, in fact, be two vertical lines through ± 4 on the x-axis. 4 4 Hence, the equation of the locus must be x = ± 4. -4 0 4 x=-4 x x=4 Loci also exist in the complex plane, of course, and are described in terms of Re z and Im z either as z = a + bi or z (x, y), or even in polar (mod-arg) form, z = rθ. 4. 𝜋 iy Sketch the graph of the locus z (x, y) if 0 ≤ arg 𝑧 ≤ 4 . 𝜋 𝜋 Arg z = 4 represents a rotation of 4 radians from the positive x-axis into the first quadrant. As the modulus is unknown, it is of indeterminate (and hence infinite) length and is best represented as a ray. 𝜋 4 0 x The locus is comprised of all the ordered pairs in the shaded area between the ray and the positive real axis, including the boundaries. Consider two points, z 1 and z 2 in the complex plane. The locus of the point z (x, y) which is equidistant from z 1 and z 2 will be the perpendicular bisector of the two points (see Examples 1 & 2 above). Consequently, if 5. |𝑧 − 𝑧1 | = |𝑧 − 𝑧2 | then z must be the perpendicular bisector of z 1 and z 2 . Sketch the graph of the locus z (x, y) if |𝑧 − 3| > |𝑧 + 2 − 𝑖|. ∴ Compare to ⇒ |𝑧 − 3| > |𝑧 + 2 − 𝑖| |𝑧 − 3| > |𝑧 − (−2 + 𝑖)| |𝑧 − 𝑧1 | = |𝑧 − 𝑧2 | z 1 = (3, 0) |𝑧 − 3| = |𝑧 + 2 − 𝑖| If perpendicular bisector shown. and iy z 2 (-2, 1) z 2 = (-2, 1) then the solution is the So, for |𝑧 − 3| > |𝑧 + 2 − 𝑖|, the region to the left of the perpendicular bisector is shaded to show |𝑧 − 3| is greater than the other side as shown in the next diagram. z (x, y) z 1 (3, 0) 0 x iy z (x, y) |𝑧 − 3| > |𝑧 + 2 − 𝑖| z 2 (-2, 1) z 1 (3, 0) 0 EXERCISE New QMaths 12C page 142, Ex 6.1 x