Coordinate Geometry Locus 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 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)

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. 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.

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.

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.

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.

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.