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## Educational Communications and Technology

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**EDD 5161F**Educational Communications and Technology**Pre-requisite Knowledge**• Distance Formula • Slope • Straight Line Drawing**Contents**Review:Distance and Slope Points of Division Equation of Straight Lines Perpendicular and Parallel Lines Intersection of Two Straight Lines**Equation of Special Lines**Two Point Form Point-Slope Form Slope-Intercept Form Intercept Form General Form Back =>**Topic: Distance and Slope**(Review) Back=>**Distance**P (x1, y1) Q (x2,y2) y1 y2 x2 x1 y1 – y2 x1 – x2 distance ? By Pythagoras Theorem,**Example**A = (-4, 2) B=(2, -4) (x1,y1) (x2,y2)**Class Work**(a) Find AB if A=(4,0) and B=(9,a) (Give the answer in terms of a.) (b) If AB= , find a. (b) (a) 2 min 2 min**Slope**P (x1, y1) Q (x2,y2) y1 – y2 x1 – x2 slope ?**Example**(4, 5) (x1,y1) (1, 1) (x2,y2)**Class Work**B(a2, ab) 3 min Slope of AB? A(b2, –ab)**Example**(2,2) (-4, 2) If a line//x-axis slope = 0**Example**(2,2) zero! (2,-3) If a line // y-axis slope is undefined**End of Topic**Back=>**Topic: Point of Division**Back=>**y**B(8,9) 9 (9-y) Point C? 8 4 7 6 A= (1,2), B = (8,9) and AC : CB = 3 : 4 Find C(x,y) C(x,y) (8-x) 5 (y -2) 4 3 3 (x-1) 2 A(1,2) 1 x -2 -1 1 2 3 4 5 6 7 8**B(8,9)**(9-y) 4 C(x,y) D (8-x) C(x,y) 3 (y -2) A(1,2) E (x-1) ∵ ΔBCD ～ΔCAE**Observation**B(8,9) Calculation Section Formula A(1,2) 4 3 C(x,y) 3 x 8 + 4 x 1 x = + 3 4 3 x 9 + 4 x 2 y = + 4 3**Example**2 P (a, b) 1 A (1, 2) B (4, 8) What are the coordinates of P ? Ans: P = (2, 4)**ClassWork**2 P (3, 1) 5 B (4, 9) A (a, b) 5 min Find the values of a and b**Class Work**1 P (a, b) 1 A (3, -7) B (5, 3) 3 min Find the coordinates of point P**1**P (a, b) 1 A(x1,y1) B (x2,y2) Mid-Point Formula P is the mid-point of AB**Example**Let P = (a, b) & G = (p, q) A (3, 4) P 2 1 2 G 1 1 2 C (-2, -5) (4, 1) B (5, -2)**Observation**A (x1,y1) G(x,y) B (x2,y2) Then: C (x3, y3) Given : G is the centroid of △ABC**End of Topic**Back=>**Challenge**A (-3, 4) P B (2, 1) Find the ratio of AP : PB. Answer =>**Solution**Let AP : PB = 1 : k Back =>**Topic: Equations of**Special Lines Back=>**Horizontal Lines**(-3, 3) (3, 3) (2, -3) (-5, -3) x = -3 (1, 3) y = 3 (2,1) y = 1 y = -3 (-1, -3)**Vertical Lines**(-1, -4) x = 2 x = -3 x = -1 (2, 2) (2, 0) (-3, -3) x = -3**Class Work**y x 0 3 min • Find • The equations ofL1 and L2; • The coordinatesof point P. (a, b) L2 P • Ans: • L1 : x = a L2 : y = b • P= (0, b) L1**y =-2x**Straight lines Passing through origin (3,3) (-1,2) (1,1) (1,-2) (-2,4) (-3,-3) y = x**Observation**(a,b)**Class Work**y L2 (6,7) (4,-3) Ans: L1 : L2 : x 3 min L1 Find the equations of L1 and L2.**End of Topic**Back=>**Topic: Two-Point Form**Back=>**L**B(5, 8) P(x, y) A(1, 3) MAP = MAB Find the equation of L.**Will the result be the same if We consider MBP instead of**MAP ? L: 5x-4y+7=0 B(5, 8) P(x, y) A(1, 3) MBP = MAB**Example**L (-4, 4) (-2, b) P (2, -3) (a) Find the equation of L. (c) Find the coordinates of P. (b) Find the value of b. L: 7x + 6y + 4 = 0**Example**The point (7, -4) lies on the straight line. The point (3, -2) does not lie on the line. (a) Find the equation of the straight line joining (-3, 2) and (2, -1). L: 3x - 5y + 1 = 0 (b) Does the point (7, -4) lie on the straight line ? (c) State whether the point (3, -2) lies on the straight line or not.**Class Work**Solution. (a) (b) 7 min (a) Find the equation of the straight line which passes through (0,0) and (-4,-6). (b) If the point A(a,3) lies on L, find a.**End of Topic**Back=>**Topic: Point-Slope Form**Back=>**A(1, 7)**L B(x, y) slope = 3 Point-slope Form MAB = Slope**Example**Find the equation of the line which passes through (-1,-5) and has slope -3 : Working? Solution**Example**(a) Find the equation of L. (b) What is the value of b ? (-3, 2) L B (2, b) Put B(2, b) into the equation L: x + 3y - 3 = 0**Class Work**Find (a) The equation of L. (b) The coordinates of P (c) The coordinates of Q L P (a, 2) (-2, 0) Q 10 min