Intersection of Straight lines Objectives To be able
Intersection of Straight lines Objectives ─ ─ To be able to find the intersection Pt of 2 lines To solve simultaneous equations To solve the intersection of a curve and a line To Sketch some specific types of curve You should know how to find the equation of a line (from the previous lesson) * INTO Foundation L 6 MH
Find the area of the triangle ABC A C B How do we do this ? ? 1. Find which line is which 2. Solve equations for ABC 3. Find Perpendicular line to BC through A * INTO Foundation L 6 MH
Solve for coordinates of A, B, C Pt A Intersection of y = 2 x + 5 1. y = -2 x + 6 2. 2 y= 11 Add to remove x y= 5. 5 Þ x=(5. 5 – 5)/2 Subst y into 1. Þ x= 0. 25 A = (0. 25, 5. 5) * INTO Foundation L 6 MH
Solve for coordinates of A, B, C Pt B Intersection of y = -0. 75 x + 2. 5 1. y = -2 x +6 2. 0 = 1. 25 x -3. 5 Subtract to remove y 3. 5= 1. 25 x Þ x=3. 5/1. 25 Þ x= 2. 80 Subst x into 2. (y=-5. 6+6) Þ y= 0. 40 B = (2. 80, 0. 40) * INTO Foundation L 6 MH
Solve for coordinates of A, B, C Pt C Intersection of y = -0. 75 x + 2. 5 1. y = 2 x +5 2. 0 = -2. 75 x -2. 5 Subtract to remove y 2. 5= -2. 75 x Þ x=2. 5/-2. 75 Þ x= -0. 9090 (~-10/11) Subst x into 1. (y=30/44+2. 5) Þ y= 3. 8181 (378/99 or 42/11) C = (-10/11, 42/11) * INTO Foundation L 6 MH
Found A, B, C A (0. 25, 0. 55) C(-10/11, 43/11) B(2. 80, 0. 40) Find the line that is perpendicular to BC through the point A * INTO Foundation L 6 MH
Find Pt D y = 4/3 x +0. 217 Grad BC is -0. 75 Grad AD is +4/3 m 1 m 2=-1 D AD is y = 4/3 x +c through A is 0. 55=4/3(1/4)+c => c=0. 216667 * INTO Foundation L 6 MH
Find Pt D Pt C Intersection of y = -0. 75 x + 2. 50 y = 1. 33 + 2. 17 1. 2. To eliminate mult eqn 1. x 1. 33 eqn 2. x 0. 75 1. 33 y = -0. 9975 x + 3. 3250 0. 75 y = 0. 9975 x + 1. 6275 Þ Þ Þ * 1. 2. D= (0. 157, 2. 38) 2. 08 y = 4. 9525 y= 4. 9525/2. 08 = 2. 38 Subst into 2. x = (0. 75(2. 08) -0. 16275)/0. 9975 = 0. 1579 INTO Foundation L 6 MH
Now Find Pt D Area For a triangle this is ½ x base x Perpendicular height So ½ x BC x AD A ≡ (0. 25, 5. 5) C = (-10/11, 42/11) B(2. 80, 0. 40) D= (0. 157, 2. 38) Length of BC is [(-10/11 -2. 8)2 + (42/11 -0. 40)2]1/2 = 5. 04 Length of AD is [(2. 38 -5. 5)2 + (0. 157 -0. 25)2] = 3. 12 So Area is 0. 50 x 5. 04 x 3. 12 = 7. 9 m 2 (1 dec pl) * INTO Foundation L 6 MH
Exercises Do worksheet If you finish do the following Plot on graph paper on the same graph the curves : 1. 2. 3. 4. 5. * y = x 2 y = x 3 y = 1/x 2 y = x 4 INTO Foundation L 6 MH
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