1. 1 Press Ctrl-A ©G Dear 2010 – Not to be sold/Free to use IntersectingLines Stage 6 - Year 11 Applied Mathematic (Preliminary Extension 1)

2. 2 End of Slide Intersecting Lines meet at P(x,y) Press Concurrent Lines meet at a point P(x,y) y P(x,y) x To find the point P(x,y) we solve the equations simultaneously.

3. 3 P(x,y) End of Slide y x Simultaneous Equation - Substitution y = 3x + 2 y - 2x - 4 = 0 (1) (2) Substitute (1) into (2). (3x +2) - 2x - 4 = 0 x - 2 = 0 x = 2x = 2 Substitute x = 2 into (1). y = 3(2) + 2 y = 8y = 8 Solution (2,8) Press P(2,8)

4. 4 P(x,y) End of Slide y x Simultaneous Equation – Elimination (1/2) x – 2y – 5 = 0 x – 3y - 2 = 0 (1) (2) (1) - (2)y - 3 = 0 y= 3 Substitute y = 3 into (1). x - 2(3) – 5 = 0 x - 11 = 0 Solution (3,11) Press P(11,3) x= 11

5. 5 -6x + 9y + 3 = 0 P(x,y) End of Slide y x Simultaneous Equation – Elimination (2/2) 3x – 2y – 4 = 0 -2x + 3y + 1 = 0 (1) (2) (1) x 2 Solution (1,2) Press P(2,1) 6x – 4y – 8 = 0 (2) x 3 (1) + (2)5y - 5 = 0 5y = 5 y = 1y = 1 Sub x=1 in (1) 3x – 2(1) – 4 = 0 3x – 6 = 0 3x = 6 x= 2

6. 6 End of Slide y x Equation of a line through the intersection of two other lines. a 2 x + b 2 y +c 2 = 0 Press a 1 x + b 1 y +c 1 = 0 (a 1 x + b 1 y + c 1 ) + k(a 2 x + b 2 y + c 2 ) = 0 P(x 1,y 1 ) Proof P(x 1,y 1 ) satisfies both equations.  (a 1 x + b 1 y + c 1 ) = 0 & (a 2 x + b 2 y + c 2 ) = 0  P(x 1,y 1 ) satisfies  (a 1 x + b 1 y + c 1 ) + k(a 2 x + b 2 y + c 2 ) = 0 0 + k(0) = 0 Sub in (a 1 x + b 1 y + c 1 ) + k(a 2 x + b 2 y + c 2 ) = 0 so passes through P(x 1,y 1 ).

7. 7 End of Slide y Example: Find the equation of the line through and the point (1,2). a 2 x + b 2 y +c 2 = 0 Press a 1 x + b 1 y +c 1 = 0 3x – 2y - 4 = 0, -2x + 3y + 1 = 0 P(x 1,y 1 ) (3x – 2y – 4) + k(-2x + 3y + 1) = 0 (a 1 x + b 1 y + c 1 ) + k(a 2 x + b 2 y + c 2 ) = 0 Find k by substituting x = 1 and y = 2 (3(1) – 2(2) – 4) + k(-2(1) + 3(2) + 1) = 0 -5 + 5k= 0 5k= 5 k= 1 (3x – 2y – 4) + k(-2x + 3y + 1) = 0 3x – 2y – 4 + 1(-2x + 3y + 1) = 0 3x – 2y – 4 -2x + 3y + 1 = 0 x + y – 3 = 0 P(1,2)