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