1. Coordinate Geometry Locus I
By Mr Porter

2. Definition: A locus is a set of points in a plane that satisfies some geometric condition or some algebraic equation. A locus is the ‘path traced out by a particle moving in a plane’ and a Cartesian equation gives us the name of the curve along which the particle travels. Very much like a jets vapour trial through the blue sky.
Loci that we have used and should already know:
Straight Lines
Horizontal Lines:
y = b
Vertical Lines:
x = a
Sloping Lines:
y = mx + b or
ax + by + c = 0
Parabolas (relate to Quadratics)
y = ax2 + bx + c, a ≠0
a < 0 concave down
a > 0 concave up
Circles, with radius r
x2 + y2 = r2, centred (0,0)
(x – h)2 + (y – k)2 = r2 , centred (h,k)

3. Assumed Knowledge: Student should be familiar with the following coordinate geometry formula:
Assumed Knowledge: Student should be familiar with the following coordinate geometry condition for:
Parallel lines: m1 = m2
Perpendicular lines: m1 . m2 = -1
Assumed Notation: Student should be familiar with the following geometry notation:
We will use the Line Interval to represent all lines.

4. Example 1: Find the locus of a point P(x, y) such that its distance from A(1,2) is equal to it distance from B(5,8).
Hint: Try to sketch a rough diagram of the information.
Hint: Interval AP equals Interval BP.
A(1,2)
B(5,8)
P(x,y) d1 d2
Then, dPA = dPB
Hint: Square both side to remove Square Root sign, √.
Hint: Expand brackets, use distributive law or F.O.I.L
Hint: Simply, rearrange for = 0.
Hint: Most of these questions are solve by the distance, gradient or midpoint coordinate geometry formulae. In this case, the distance formula.
Hint: Reduce, divide by 4
This the general from of a (sloping) line, of the form ax + by + c = 0.

5. Example 2: Find the locus of a point P(x, y) such that its distance from A(-5,2) is equal to it distance from B(4,-3).
Hint: Try to sketch a rough diagram of the information.
Hint: Interval AP equals Interval BP.
A(-5,2)
B(4,-3)
P(x,y) d1 d2
Then, dPA = dPB
Hint: Square both side to remove Square Root sign, √.
Hint: Expand brackets, use distributive law or F.O.I.L
Hint: Simply, rearrange for = 0.
Hint: Most of these questions are solve by the distance, gradient or midpoint coordinate geometry formulae. In this case, the distance formula.
Hint: Reduce, divide by 2
This the general from of a (sloping) line, of the form ax + by + c = 0.

6. This the general from of a vertical straight line, of the form x = a.
Example 3: What is the locus of a point P(x, y) that is always 3 units from the line x = 5.
Hint: Try to sketch a rough diagram of the information.
Hint: Interval PM equals 3 units. 5
Line: x = 5
P(x,y)
d = 3
Then, dPM = 3
M(5,y)
Hint: Square both side to remove Square Root sign, √.
Hint: Expand brackets, use distributive law or F.O.I.L
Hint: Simply, rearrange for = 0.
Hint: The shortest distance from P(x,y) to the line x = 5 is the perpendicular distance.
This distance is horizontal to the line would intersect at M. ‘M’ must have the same y-coordinate as P(x,y) and its x-coordinate has to be x = 5 to lie on the line.
Now, using the distance formula:
Hint: Factorise.
Hint: Solve for x.
This the general from of a vertical straight line, of the form x = a.
This represent 2 vertical lines.
Note: There are other LOGICAL ways of solving this problem.
Such as using the perpendicular distance formula.

7. This the general from of a straight line, of the form ax + by +c =0.
Example 4: What is the locus of a point P(x, y) which is √2 units from the line y = x – 1.
Hint: Try to sketch a rough diagram of the information.
Hint: Interval PM equals √2 units.
P(x,y) -1
Line: y = x
d = √2 M
Line: Ax + By + C = 0
i.e. x – y = 0  A =1, B = -1, C = 0
(x,y) are coordinates of P.
Substituting values and removing absolute sign (replace with ±, LHS)
Evaluate LHS.
Rearrange, take care with ±.
Hint: The shortest distance from P(x,y) to the line y = x – 1 is the perpendicular distance.
Using the perpendicular distance formula from a point to a line:
Write separate equations in general form.
This the general from of a straight line, of the form ax + by +c =0.
Note: There are other LOGICAL ways of solving this problem.

8. This the general from of a vertical straight line.
Example 5: A(a,0) and B(0,-a), find the locus of P(x,y) such that the gradient AP is twice the gradient of BP.
Hint: Try to sketch a rough diagram of the information.
P(x,y)
A(a,0)
B(-a,0)
Evaluate LHS and RHS.
Hint: There we must use the gradient formula fro 2 points:
m AP
m BP
Rearrange, by cross multiplying.
Expand, simplify and rearrange.
Divide by y.
This the general from of a vertical straight line.