2 3 2 3 6 3 2 6 2 x y y x y x The slope is 2 3 Since the lines are parallel 2 3

# 2 3 2 3 6 3 2 6 2 x y y x y x the slope is 2 3 since

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2 3 2 3 6 3 2 6 2 x y y x y x + = = − + = − + The slope is 2 3 . Since the lines are parallel, 2 3 is also the slope of the line whose equation is to be found. Substitute 2 3 m = − , 1 3 x = − , and 1 2 y = into the point-slope form. ( ) ( ) ( ) ( ) 2 1 1 3 2 3 2 3 3 2 2 3 3 6 2 – 6 3 –2 y y m x x y x y x y x y x y x = = − − − = − + = − = = − 16. the vertical line through ( ) 4,3 The equation of a vertical line has an equation of the form x = a . Since the line passes through ( ) 4,3 , the equation is x = 4. 17. 2 2 4 2 4 x x y y + + = Complete the square on x and y separately. ( ) ( ) ( ) ( ) ( ) ( ) 2 2 2 2 2 2 4 2 4 4 4 2 1 4 4 1 2 1 9 x x y y x x y y x y + + = + + + + = + + + + = Yes, it is a circle. The circle has its center at ( ) 2, 1 and radius 3.
Summary Exercises on Graphs, Circles, Functions, and Equations 205 18. 2 2 6 10 36 0 x x y y + + + + = Complete the square on x and y separately. ( ) ( ) ( ) ( ) ( ) ( ) 2 2 2 2 2 2 6 10 36 6 9 10 25 –36 9 25 3 5 2 x x y y x x y y x y + + + = − + + + + + = + + + + + = − No, it is not a circle. 19. 2 2 12 20 0 x x y + + = Complete the square on x and y separately. ( ) ( ) ( ) 2 2 2 2 2 2 12 20 12 36 –20 36 6 16 x x y x x y x y + = − + + = + + = Yes, it is a circle. The circle has its center at (6, 0) and radius 4. 20. 2 2 2 16 61 x x y y + + + = − Complete the square on x and y separately. ( ) ( ) ( ) ( ) ( ) ( ) 2 2 2 2 2 2 2 16 61 2 1 16 64 –61 1 64 1 8 4 x x y y x x y y x y + + + = − + + + + + = + + + + + = Yes, it is a circle. The circle has its center at ( ) 1, 8 and radius 2. 21. 2 2 2 10 0 x x y + + = Complete the square on x and y separately. ( ) ( ) ( ) 2 2 2 2 2 2 2 10 2 1 –10 1 1 9 x x y x x y x y + = − + + = + + = − No, it is not a circle. 22. 2 2 – 8 9 0 x y y + = Complete the square on x and y separately. ( ) ( ) ( ) 2 2 2 2 2 2 – 8 9 – 8 16 9 16 – 4 25 x y y x y y x y + = + + = + + = Yes, it is a circle. The circle has its center at ( ) 0,4 and radius 5. 23. The equation of the circle is 2 2 2 ( 4) ( 5) 4 x y + = . Let y = 2 and solve for x : 2 2 2 ( 4) (2 5) 4 x + = 2 2 2 2 ( 4) ( 3) 4 ( 4) 7 x x + − = = 4 7 4 7 x x = ± = ± The points of intersection are ( ) 4 7,2 + and ( ) 4 7,2 24. Write the equation in center-radius form by completing the square on x and y separately: ( ) ( ) 2 2 2 2 2 2 2 2 10 24 144 0 10 24 144 0 ( 10 25) ( 24 144) 25 ( 5) ( 12) 25 x y x y x x y y x x y y x y + + = + + + = + + + = + = The center of the circle is (5, 12) and the radius is 5. Now use the distance formula to find the distance from the center (5, 12) to the origin: ( ) ( ) 2 2 2 1 2 1 2 2 (5 0) (12 0) 25 144 169 13 d x x y y = + = + = + = = Since the radius is 5, the shortest distance from the origin to the graph of the circle is 13 - 5 = 8. 25. (a) The equation can be rewritten as 6 3 1 1 4 4 4 2 4 6 . y x y x y x = − = + = + x can be any real number, so the domain is all real numbers and the range is also all real numbers. domain: ( ) , ; −∞ ∞ range: ( ) , −∞ ∞ (b) Each value of x corresponds to just one value of y . 4 6 x y = − represents a function.
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