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Applied Mathematic (Preliminary Extension 1) Intersecting Lines Stage 6 - Year 11 Press Ctrl-A ©G Dear 2010 – Not to be sold/Free to use 1 Concurrent Lines Intersecting Lines meet point meetat ataP(x,y) P(x,y) y To find the point P(x,y) we solve the equations simultaneously. P(x,y) x Press End of Slide 2 Simultaneous Equation - Substitution y = 3x + 2 y - 2x - 4 = 0 y (1) (2) Substitute (1) into (2). P(x,y) P(2,8) (3x +2) - 2x - 4 = 0 x-2=0 x=2 Substitute x = 2 into (1). x Press Solution (2,8) y = 3(2) + 2 y=8 End of Slide 3 Simultaneous Equation – Elimination (1/2) x – 2y – 5 = 0 x – 3y - 2 = 0 y (1) (2) (1) - (2) P(x,y) P(11,3) y -3=0 y= 3 Substitute y = 3 into (1). x - 2(3) – 5 = 0 x - 11 = 0 x x= 11 Press Solution (3,11) End of Slide 4 Simultaneous Equation – Elimination (2/2) 3x – 2y – 4 = 0 -2x + 3y + 1 = 0 y P(x,y) P(2,1) (1) x 2 (2) x 3 (1) (2) 6x – 4y – 8 = 0 -6x + 9y + 3 = 0 (1) + (2) Press 5y - 5 = 0 5y = 5 Sub x=1 in (1) y=1 3x – 2(1) – 4 = 0 x 3x – 6 = 0 3x = 6 Solution (1,2) x= 2 End of Slide 5 Equation of a line through the intersection of two other lines. (a1x + b1y + c1) + k(a2x + b2y + c2) = 0 y Proof P(x1,y1) satisfies both equations. P(x1,y1) Press (a1x + b1y + c1) = 0 & (a2x + b2y + c2) = 0 Sub in (a1x + b1y + c1) + k(a2x + b2y + c2) = 0 x 0 + k(0) = 0 P(x1,y1) satisfies (a1x + b1y + c1) + k(a2x + b2y + c2) = 0 so passes through P(x1,y1). End of Slide 6 Example: Find the equation of the line through 3x – 2y - 4 = 0, -2x + 3y + 1 = 0 and the point (1,2). (a1x + b1y + c1) + k(a2x + b2y + c2) = 0 y (3x – 2y – 4) + k(-2x + 3y + 1) = 0 P(x1,y1) P(1,2) Press Find k by substituting x = 1 and y = 2 (3(1) – 2(2) – 4) + k(-2(1) + 3(2) + 1) = 0 -5 + 5k= 0 (3x – 2y – 4) + k(-2x + 3y + 1) = 0 5k= 5 3x – 2y – 4 + 1(-2x + 3y + 1) = 0 3x – 2y – 4 -2x + 3y + 1 = 0 k= 1 x+y– 3 =0 End of Slide 7