This preview shows page 1. Sign up to view the full content.
Unformatted text preview: MAC1105: Quiz #15 Solutions
July 19, 2010 In the topright 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 slopeintercept form (Hint: Find the pointslope 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 pointslope 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 xintercept and y intercept of the line 4x  5y = 7, and then nd its slope. To get the xintercept, we plug in y = 0 and solve for x: 4x = 7 x = 7 . 4 So the xintercept 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 ...
View Full
Document
 Summer '10
 Picklesimer
 Algebra

Click to edit the document details