Self Test Exercise 8 Solution 9 6 1 P 23 6 x y Rearranging to slope intercept

# Self test exercise 8 solution 9 6 1 p 23 6 x y

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Self-Test Exercise 8, Solution 9 –6, –1 P 2 3 6 x y Rearranging to slope-intercept form, 3 2 6 y x  Dividing both sides by 3, 2 2 3 y x  The slope of the line parallel to this will have the same slope, 2 3 m  y mx b Substituting the slope and the coordinates –6, –1 P 2 -1 (–6) 3 b  -1 4 b   1 4 b 5 b  Therefore, the equation of the line that is parallel to 2 3 6 x y that passes through –6, –1 P is 2 3 5 y x  or 2 3 5 1 x y  Self-Test Exercise 8, Solution 11 (-3, 5) and (5, -1) 2 1 2 1 = y y x m x 6 = = ( 3) 5 8 1 5 m y = mx + b 3 y = - x + b 4 Substitute (-3, 5) to the equation, 3 = - (-3) 5 + b 4 9 = + b 4 5 9 - = b 4 5 11 b = 4 3 11 y = - x + 4 4 3 11 4(y = - x + ) 4 4 Multiply both sides by 4, 4y = -3x + 11 Convert to standard form, 3x+4y = 11 Self-Test Exercise 8, Solution 13 0,0 P 5 1 y x Rearranging to slope-intercept form, The slope of the line parallel to this will have the same slope, 5 m y mx b Substituting the slope and the coordinates 0,0 P 5 y x b 0 5(0) b 0 b Therefore, the equation of the line that is parallel to 5 1 y x that passes through 0,0 P is 5 y x or 5 0 x y Self-Test Exercise 8, Solution 15 4 3 16 0 x y 3 -4 16 y x 3 4 16 - 3 3 3 y x 4 16 - 3 3 y x 4 3 16 3 m b  2 2 0 x y 2 2 y x 2 2 m b The slopes of the lines are different. Therefore, the system has one solution. Self-Test Exercise 8, Solution 17 3 6 y x 3 6 m b 6 – 2 12 0 x y 6 12 2y x 2y 6 12 x 2 6 12 y 2 2 2 x y 3 6 x 3 6 m b The slope of the lines and y-intercept are the same. Therefore, the system has infinite many solutions.  #### You've reached the end of your free preview.

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