1. 1 Press Ctrl-A ©G Dear2008 – Not to be sold/Free to use Finding the Equation Stage 6 - Year 11 Mathematic (Preliminary)

2. 2 A locus describes a set of points P(x,y) that obeys certain conditions, or a single point P(x,y) that moves along a certain path. Locus

3. 3 E.g. 1 - Find the equation of the locus of a point P(x,y) that moves so that it is always 3 units from the origin. Remember x 2 + y 2 = r 2 Distance Formula d 2 = (x 2 – x 1 ) 2 + (y 2 – y 1 ) 2 3 2 = (x – 0) 2 + (y – 0) 2 From origin (0,0) to point (x, y) x 2 + y 2 = 9Locus

4. 4 P E.g. 2 - Find the equation of the locus of a point that is always equidistant from two fixed points A(1, -2) and B(3, 4). Locus AP = BP  AP 2 = BP 2 (x-1) 2 + (y--2) 2 = (x-3) 2 + (y-4) 2 (x-1) 2 + (y+2) 2 = (x-3) 2 + (y-4) 2 x 2 –2x+1 + y 2 +4y+4 = x 2 –6x+9 + y 2 –8y+16 4x + 12y – 20 = 0 x + 3y – 5 = 0Locus

5. 5 E.g. 3 - Find the equation of the locus of a point P(x,y) that moves so that the distance PA to distance PB is in the ratio 2:1 where A = (-3, 1) and B = (2, -2). Locus PA:PB =2:1 ---- = --- 2121 PA PB PA = 2PB PA 2 = (2PB) 2 = 4PB 2 (2, -2) (-3, 1)

6. 6 E.g. 3 - Find the equation of the locus of a point P(x,y) that moves so that the distance PA to distance PB is in the ratio 2:1 where A = (-3, 1) and B = (2, -2). Locus i.e. (x––3) 2 + (y–1) 2 = 4[(x-2) 2 + (y+2) 2 ] Use d 2 = (x 2 –x 1 ) 2 + (y 2 –y 1 ) 2 (x+3) 2 + (y–1) 2 = 4[(x-2) 2 + (y+2) 2 ] x 2 +6x+9 + y 2 –2y+1 = 4(x 2 –4x+4 + y 2 +4y+4) x 2 +6x+9 + y 2 –2y+1 = 4x 2 –16x+16 + 4y 2 +16y+16 0 = 3x 2 –22x+3y 2 +18y+22 3x 2 –22x+3y 2 +18y+22 = 0 PA 2 :PB 2 =4:14:1

7. 7 E.g. 4 - Find the equation of the locus of a point P(x,y) that moves so that point is equidistant from a fixed point A = (1, -2) and fixed line y=5. Locus PA = PB PA 2 = PB 2 (x–x) 2 + (y–5) 2 = (x –1) 2 + (y -- 2) 2 (x–x) 2 + (y–5) 2 = (x –1) 2 + (y+2) 2 y 2 -10y+25 = x 2 –2x+1 + y 2 +4y+4 0 = x 2 –2x+5y+14y–20 P x 2 –2x+5y+14y–20 = 0

8. 8 E.g. 5 - Find the equation of the locus of a point P(x,y) that moves so that the line PA is perpendicular to PB, where A = (1, 2) and B = (-3, -1). Locus For perpendicular lines m 1 m 2 =-1 PA: m 1 =PB: m 2 = For PA perpendicular to PB. m1m1 y 2 – y – 2 = –(x 2 + 2x – 3) y 2 – y – 2 = –x 2 – 2x + 3 x 2 +2x+y 2 –y–5 = 0 P y – –1 x – – 3 = y + 1 x + 3 y – 2 x – 1 = -1 y + 1 x + 3 y – 2 x – 1 x x m 2 = -1 y 2 –y – 2 x 2 + 2x – 3 = -1 Using m = y 2 – y 1 x 2 – x 1

9. 9 E.g. 6 - Find the equation of the locus of a point P(x,y) that moves so that its distance from the x-axis is 3 times its distance from the y-axes. Locus PA = 3PB PA 2 = (3PB) 2 PA 2 = 9PB 2 (x–x) 2 + (y–0) 2 = 9[(x–0) 2 + (y–y) 2 ] y 2 = 9x 2 y = ±3x

10. 10 Find the equation of the locus of a point moveing so that its distance from the line 3x+4y+5=0 is always 4 units. Locus P1 P2 Since lines are parallel (m1 = m2) then we just need to find the constant ‘c’ in 3x+4y+c=0. Use d = |ax + by + c| a 2 + b 2 4 = |3 x 0 + 4 x -1.25 + c| 3 2 + 4 2 a = 3b = 4 x = 0y = -1.25 4 = |c - 5| 5 20 = |c - 5| c – 5 = +20 c – 5 = -20 c = +25 c = -15  3x + 4y + 25 = 0  3x + 4y - 15 = 0 If x = 0 then 4y + 5 = 0 4y = -5 y = -1.25