Quiz15solutions - MAC1105: Quiz #15 Solutions July 19, 2010...

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Unformatted text preview: MAC1105: Quiz #15 Solutions July 19, 2010 In the top-right corner of a clean sheet of paper, write your name, UFID, and section number. Please use a pen with blue or black ink. When you are nished, FOLD your paper in half lengthwise and write your name on the back. 1. Write the equation for the line passing through the points (-6,-4) and (-3,3) in slope-intercept form (Hint: Find the point-slope form rst). 3-(-4) y2 -y 3+4 To nd the slope, we set up m = x2 -x1 = -3-(-6) = -3+6 = 7 . So with 3 1 point-slope form, we have y - y1 = m(x - x1 ) y - (-6) = 7 [x - (-4)] 3 7 y = 7 x + 28 - 6 y = 3 x + 10 . 3 3 3 2. Find the x-intercept and y -intercept of the line 4x - 5y = 7, and then nd its slope. To get the x-intercept, we plug in y = 0 and solve for x: 4x = 7 x = 7 . 4 So the x-intercept is ( 7 , 0). To get the y -intercept, we plug in x = 0 and 4 7 solve for y : -5y = 7 y = - 7 . So the y -intercept is (0, - 5 ). The slope 5 7 - 5 -0 y is then m = x2 -y1 = 0- 7 = 4 . 5 2 -x1 4 3. Find the equation of the line perpendicular to the line -x + 2y = 4 which passes through the origin. We need the slope of the given line. Solving for y gives y = 1 (x+4), so the 2 slope is 1 . To get the slope perpendicular, we take the negative reciprocal 2 of 1 , which is -2. Since it passes through the origin, the equation is 2 simply y = -2x. 4. Find the equation of the line parallel to the line x = 2 passing through the point (5, 3). Since x = 2 is a vertical line, the only line that will work is x = 5. 1 ...
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