Year 12 – C1 Straight Line Graphs.

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Presentation transcript:

Year 12 – C1 Straight Line Graphs

A θ B D C If two lines are parallel they have equal gradients. If two lines are perpendicular the product of their gradients is -1. Proof 2 1 A θ B D C Grad of 1 = AD/DB = tanθ Grad of 2 =-AD/CD= -1/ tanθ tanθ x -1/ tanθ = -1

2, -½ 2/3, -3 -8, ¼ 13/5, -5/8 -3/4 because 4/3 x -3/4 = -1 Which of these pairs of gradients apply to pairs of perpendicular lines? 2, -½ 2/3, -3 -8, ¼ 13/5, -5/8 What would the gradient of a line be if it was perpendicular to another line with gradient 4/3? -3/4 because 4/3 x -3/4 = -1

Example Given the points A(1, -1), B(5, 2), P(-1, 10), Q(-1, 3) and R(-1, -3) show that AB is parallel to QR and that BP is perpendicular to AB. Grad AB = 2 - -1 Grad QR = -3 – 3 Grad BP = 10 - 2 5 – 1 -9 - -1 -1 – 5 = ¾ = ¾ = -4/3 Therefore AB and QR parallel, BP perpendicular to AB.

y = mx + c m = gradient c = y intercept Find the gradients and y intercepts of these lines. 2y = 3x – 4 x + 4y = 2

Example A(8, 2) lies on the line x – 2y – 4 = 0. Find the equation of the straight line parallel to x – 2y – 4 = 0 that passes through B(6, 6). b) Find the equation of the line perpendicular to x – 2y – 4 = 0 that passes through B(6, 6). a) Rearranging equation: y = ½x – 2 Grad = ½ m = y – y1 ½ = y – 6 y = ½x + 3 x – x1 x – 6 b) If perpendicular Grad = -2 m = y – y1 -2 = y – 6 y = -2x + 18 x – x1 x - 6

Equation of a line joining 2 points y – y1 = y2 – y1 x – x1 x2 – x1 Example Find the equation of line passing through (3, -1) and (7, 2) y - -1 = 2 - -1 y + 1 = ¾ x – 3 7 – 3 x - 3 y + 1 = ¾(x – 3) y = ¾ x – 9/4 – 1 y = ¾ x – 5/4

Further Example Find the equation of the perpendicular bisector of the line joining (2, -3) and (6, 5) Mid point = (2 + 6, -3 + 5) = (4, 1) 2 2 Gradient = 5 - -3 = 2 6 – 2 Grad of Perp Bis = - ½ m = y – y1 -1/2 = y – 1 - ½(x – 4) = y - 1 x – x1 x - 4 y = -1/2 x + 3

Exercise Find the equation of the line that: passes through (-1, -6) with gradient 5 passes through (4, 1) parallel to 2y=x – 3 passes through (-2,9) perpendicular to 2y = x – 3 d) passes through (0,4) and (3, 10) e) passes through (0,7) and (7,0) f) passes through (6,-2) and (12, 1) 2. Find the equation of the perpendicular bisector of the line joining the points (4,1) and (2,7) 3. A triangle has vertices at A(0,7) B(9,4) and C(1,0). Find The equation of the perpendicular from C to AB The equation of the straight line from A to the mid point of BC.