Ch.6_Solution_Manual_Ed.1_v4_

Ch.6_Solution_Manual_Ed.1_v4_ - Exercises 6.1 Exercise 6.1,...

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Exercises 6.1 Exercise 6.1, Solution 1: Exercise 6.1, Solution 3: a. A lies in the second quadrant b. B lies in the fourth quadrant c. C lies in the first quadrant d. D lies on the X -axis e. E lies in the third quadrant f. F lies on the Y -axis Exercise 6.1, Solution 5: a. The length of the horizontal line passing between these two points will be equal to the difference between their x -co-ordinates Difference between x co-ordinates = 5 – 3 = 2 units Therefore, the length of the horizontal line passing between (3, 4) and (5, 4) is 2 units. b. The length of the horizontal line passing between these two points will be equal to the difference between their x -co-ordinates Difference between x co-ordinates = 2 – (7) = 2 + 7 = 9 units Therefore, the length of the horizontal line passing between (7, 1) and (2, 1) is 9 units. c. The length of the horizontal line passing between these two points will be equal to the difference between their x -co-ordinates Difference between x co-ordinates = 0 – (5) = 0 + 5 = 5 units Therefore, the length of the horizontal line passing between (5, 3) and (0, 3) is 5 units. d. The length of the horizontal line passing between these two points will be equal to the difference between their x -co-ordinates Difference between x co-ordinates = 6 – (–2) = 6 + 2 = 8 units Therefore, the length of the horizontal line passing between (–2, –2) and (6, –2) is 8 units. Exercise 6.1, Solution 7: Let the co-ordinates of the 4 th vertex, D be ( x , y ) Length of horizontal side AB = Difference between x -co-ordinates of B and A = 1 – (–3) = 4
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Length of horizontal side DC = Difference between x co-ordinates of C and D = 1 – x All sides of a square have equal lengths, therefore length of side AB = length of side DC Hence we have 4 = 1 – x 1 – x = 4 x = 1 – 4 = 3 Length of vertical side DA = Difference between y co-ordinates of A and D = 3 – y All sides of a square have equal lengths, length of side DA = 4 units Hence we have 3 – y = 4 y = 3 – 4 = – 1 Therefore, the co-ordinates of the 4 th vertex, D is (– 3, – 1) Exercise 6.1, Solution 9: The given line is a vertical line, therefore the points at the end of the line will have the same x co- ordinate, 1. Let the y co-ordinate of the point on the other side of the vertical line be ‘ y Difference between y co-ordinates of the two points will be equal to the distance of the line, 7 units. There are two possibilities in this case based on the value of ‘ y ’ compared to the y co-ordinate of the given point (1, 5): y > 5 In this case, we would have y – 5 = 7 Hence, y = 5 + 7 = 12 y < 5 In this case, we would have 5 – y = 7 Hence y = 5 – 7 = – 2 Therefore, the possible co-ordinates of the other end of the line are (1, 12) and (1, – 2) 2
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Exercises 6.2 Exercise 6.2, Solution 1: a. Let the missing value be ‘
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This note was uploaded on 04/04/2012 for the course MATH 1052 taught by Professor Kit during the Winter '12 term at Fanshawe.

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Ch.6_Solution_Manual_Ed.1_v4_ - Exercises 6.1 Exercise 6.1,...

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