1. 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) 12 October, 2015INTO Foundation L6 MH

2. Find the area of the triangle ABC 12 October, 2015INTO Foundation L6 MH How do we do this ?? 1. Find which line is which 2. Solve equations for ABC 3. Find Perpendicular line to BC through A A B C

3. Solve for coordinates of A,B,C Pt A Intersection of y = 2x + 5 1. y = -2x + 6 2. 2y= 11 Add to remove x y= 5.5  x=(5.5 – 5)/2 Subst y into 1.  x= 0.25 A = (0.25, 5.5) 12 October, 2015INTO Foundation L6 MH

4. Solve for coordinates of A,B,C Pt B Intersection of y = -0.75x + 2.5 1. y = -2x + 6 2. 0 = 1.25x -3.5 Subtract to remove y 3.5= 1.25x  x=3.5/1.25  x= 2.80 Subst x into 2. (y=-5.6+6)  y= 0.40 B = (2.80, 0.40) 12 October, 2015INTO Foundation L6 MH

5. Solve for coordinates of A,B,C Pt C Intersection of y = -0.75x + 2.5 1. y = 2x + 5 2. 0 = -2.75x -2.5 Subtract to remove y 2.5= -2.75x  x=2.5/-2.75  x= -0.9090 (~-10/11) Subst x into 1. (y=30/44+2.5)  y= (35/11) C = (-10/11, 35/11) 12 October, 2015INTO Foundation L6 MH

6. Found A,B,C 12 October, 2015INTO Foundation L6 MH A (0.25,5.5) B(2.80,0.40) C(-10/11, 35/11) Find the line that is perpendicular to BC through the point A

7. Find Pt D 12 October, 2015INTO Foundation L6 MH D Grad BC is -0.75 Grad AD is +4/3 m 1 m 2 =-1 AD is y = 4/3x +c through A is 5.5=4/3(1/4)+c => c=5.16667 y = 4/3x + 31/6 5.5 35/11

8. Find Pt D Pt C Intersection of y = -0.75x + 2.50 1. y = 1.33x + 5.17 2. To eliminate mult eqn1.x1.33 eqn 2.x0.75 1.33y = -0.9975x + 3.3250 3. 0.75y = 0.9975x + 3.8775 4. add(3+4)  2.08y = 7.2025  y= 7.7025/2.08 = 3.46 Subst into 2.  x = (3.46 – 5.15)/1.33 = -1.29 12 October, 2015INTO Foundation L6 MH D= (-1.29, 3.46)

9. Now Find Pt D Area For a triangle this is ½ x base x Perpendicular height So ½ x |BC| x |AD| C = (-10/11, 35/11) B(2.80,0.40) D= (-1.29, 3.46) Length of BC is [(-10/11-2.8) 2 + (35/11-0.40) 2 ] 1/2 = 4.64 Length of AD is [(3.46-5.5) 2 + (-1.29-0.25) 2 ] = 2.56 So Area is 0.50 x 4.64 x 2.56 = 5.9m 2 (1 dec pl) 12 October, 2015INTO Foundation L6 MH A ≡ (0.25, 5.5)

10. Exercises Do worksheet If you finish do the following Plot on graph paper on the same graph the curves : 1. y = x 2 2. y = x 3 3. y = 1/x 4. y = 1/x 2 5. y = x 4 12 October, 2015INTO Foundation L6 MH