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Pre-Calculus Homework Solutions 73

# Pre-Calculus Homework Solutions 73 - 8 x x 3y 3y y 3y 3y 3...

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8. x 3 y ¬ 6 3 y ¬ x 6 y ¬ 1 3 x 2 x 3 y ¬ 14 3 y ¬ x 14 y ¬ 1 3 x 1 3 4 First, write an equation of a line p perpendicular to the given lines. The slope of p is the opposite reciprocal of 1 3 , or 3. Use the y -intercept of y 1 3 x 2, (0, 2), as one of the endpoints of the perpendicular segment. y y 1 ¬ m ( x x 1 ) y 2 ¬ 3( x 0) y 2 ¬ 3 x y ¬ 3 x 2 Next, use a system of equations to determine the point of intersection of y 1 3 x 1 3 4 and p . 1 3 x 1 3 4 ¬ 3 x 2 1 3 x 3 x ¬ 2 1 3 4 1 3 0 x ¬ 2 3 0 x ¬ 2 y ¬ 3( 2) 2 y ¬ 4 The point of intersection is ( 2, 4). Then, use the Distance Formula to determine the distance between (0, 2) and ( 2, 4). d ¬ ( x 2 x 1 ) 2 ( y 2 y 1 ) 2 ¬ ( 2 0) 2 ( 4 2) 2 ¬ 40 ¬ 2 10 The distance between the lines is 2 10, or approximately 6.32 units. 9. 1. Graph y 3 4 x 1 4 and point P . Place the compass at point P . Make the setting wide enough so that when an arc is drawn, it intersects y 3 4 x 1 4 in two places. Label these points of intersection A and B . 2. Put the compass at point A and draw an arc below the line. 3. Using the same compass setting, put the
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