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) 31 May, 2019 INTO Foundation L6 MH

Find the area of the triangle ABC How do we do this ?? Find which line is which Solve equations for ABC Find Perpendicular line to BC through A 31 May, 2019 INTO Foundation L6 MH

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) 31 May, 2019 INTO Foundation L6 MH

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) 31 May, 2019 INTO Foundation L6 MH

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= 3.8181 (378/99 or 42/11) C = (-10/11, 42/11) 31 May, 2019 INTO Foundation L6 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 31 May, 2019 INTO Foundation L6 MH

Find Pt D D y = 4/3x +0.217 Grad BC is -0.75 Grad AD is +4/3 m1m2=-1 AD is y = 4/3x +c through A is 0.55=4/3(1/4)+c => c=0.216667 31 May, 2019 INTO Foundation L6 MH

Find Pt D Pt C Intersection of y = -0.75x + 2.50 1. y = 1.33 + 2.17 2. To eliminate mult eqn1.x1.33 eqn 2.x0.75 1.33y = -0.9975x + 3.3250 1. 0.75y = 0.9975x + 1.6275 2. 2.08y = 4.9525 y= 4.9525/2.08 = 2.38 Subst into 2. x = (0.75(2.08) -0.16275)/0.9975 = 0.1579 D= (0.157, 2.38) 31 May, 2019 INTO Foundation L6 MH

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, 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 = 7.9m2 (1 dec pl) A ≡ (0.25, 5.5) 31 May, 2019 INTO Foundation L6 MH

Exercises Do worksheet If you finish do the following Plot on graph paper on the same graph the curves : y = x2 y = x3 y = 1/x y = 1/x2 y = x4 31 May, 2019 INTO Foundation L6 MH