Coordinate GeometryLocus I

By Mr Porter

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 LinesHorizontal Lines:

y = bVertical Lines:

x = aSloping Lines:

y = mx + bor

ax + by + c = 0

Parabolas (relate to Quadratics)

y = ax2 + bx + c, a ≠0

a < 0 concave downa > 0 concave up

Circles, with radius rx2 + y2 = r2, centred (0,0)(x – h)2 + (y – k)2 = r2 , centred (h,k)

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.

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.

A(1,2)

B(5,8)

P(x,y)

d1

d2

Hint: Most of these questions are solve by the distance, gradient or midpoint coordinate geometry formulae. In this case, the distance formula.

Hint: Interval AP equals Interval BP.

Then, dPA = dPB

Hint: Square both side to remove Square Root sign, √.

Hint: Expand brackets, use distributive law or F.O.I.L

Hint: Reduce, divide by 4

This the general from of a (sloping) line, of the form ax + by + c = 0.

Hint: Simply, rearrange for = 0.

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.

A(-5,2)

B(4,-3)

P(x,y)d1

d2

Hint: Most of these questions are solve by the distance, gradient or midpoint coordinate geometry formulae. In this case, the distance formula.

Hint: Interval AP equals Interval BP.

Then, dPA = dPB

Hint: Square both side to remove Square Root sign, √.

Hint: Expand brackets, use distributive law or F.O.I.L

Hint: Reduce, divide by 2

This the general from of a (sloping) line, of the form ax + by + c = 0.

Hint: Simply, rearrange for = 0.

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: 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: Interval PM equals 3 units.

Then, dPM = 3

Hint: Square both side to remove Square Root sign, √.

Hint: Expand brackets, use distributive law or F.O.I.L

Hint: Solve for x.

Hint: Simply, rearrange for = 0.

5

Lin

e: x

= 5

P(x,y)

d = 3M(5,y)

Hint: Factorise.

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.

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: 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:

Hint: Interval PM equals √2 units.

P(x,y)

-1

Line: y

= x

d = √2

M

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.

Line: Ax + By + C = 0i.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 ±.

Write separate equations in general form.

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.

This the general from of a vertical straight line.

P(x,y)

A(a,0)B(-a,0)

Hint: There we must use the gradient formula fro 2 points:

m APm BP

Evaluate LHS and RHS.

Rearrange, by cross multiplying.

Expand, simplify and rearrange.

Divide by y.