course Mth 158 ????????????assignment #018
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20:26:29 query 2.3.34 / 30 (was 2.3.24). Slope 4/3, point (-3,2) Give the three points you found on this line and explain how you obtained them.
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RESPONSE --> If x increases by 3, y increases by 4 -3 + 3 = 0 2 +4=6 (0,6) 0 + 3 = 3 6 + 4=10 (3, 10 ) 3 +3 = 6 10 + 4 + 14 (6, 14) The three points are (0,6), (3, 10 ), (6, 14)
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20:29:23 STUDENT SOLUTION: (-3,2) slope 4/3. Move 3 units in the x direction, 4 in the y direction to get ((-3+3), (2+4)), which simplifies to (0,6) (-3,2) slope 4/3 = -4/-3 so move -3 units in the x direction and -4 in the y direction to get ((-3-3), (2-4)) which simplifies to (-6,-2) From (0,6) with slope 4/3 we move 4 units in the y direction and 3 in the x direction to get ((0+3), (6+4)), which simplifies to (3,10). The three points I obtained are (-6,-2), (0,6), (3,10).
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RESPONSE --> I moved all my points to the positive direction and did not get any negative points.
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20:44:17 query 2.3.40 / 36 (was 2.3.30). Line thru (-1,1) and (2,2) **** Give the equation of the line and explain how you found the equation.
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RESPONSE --> Slope = 2-1/2-(-1) = 1/3 Using y-y1 = m (x-x1) y-1 = 1/3 (-x-(-1) y-1 = -1/3x + 1/3 y=1/3x + 4/3
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20:44:34 STUDENT SOLUTION: The slope is m = (y2 - y1) / (x2 - x1) = (2-1)/(2- -1) = 1/3. Point-slope form gives us y - y1 = m (x - x1); using m = 1/3 and (x1, y1) = (-1, 1) we get y-1=1/3(x+1), which can be solved for y to obtain y = 1/3 x + 4/3.
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RESPONSE --> Ok
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20:50:49 **** query 2.3.54 / 46 (was 2.3.40). x-int -4, y-int 4 **** What is the equation of the line through the given points and how did you find the equation?
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RESPONSE --> x intercept = -4 y intercept = 4 (-4,0) (0,4) m= 4-0/0- (-4) = 4/4 =1 y= mx + b y=1(x) +4
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20:51:54 STUDENT SOLUTION: The two points are (0, 4) and (4, 0). The slope is therefore m=rise / run = (4-0)/(0+4) = 1. The slope-intercept form is therefore y = m x + b = 1 x + 4, simplifying to y=x+4.
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RESPONSE --> ok
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20:56:59 **** query 2.3.76 / 56 (was 2.4.48). y = 2x + 1/2. **** What are the slope and the y-intercept of your line and how did you find them?
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RESPONSE --> Slope is 2/1 and y intercept is 1/2
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20:57:16 ** the y intercept occurs where x = 0, which happens when y = 2 (0) + 1/2 or y = 1/2. So the y-intercept is (0, 1/2). The slope is m = 2.**
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RESPONSE --> ok
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21:02:00 **** query 2.3.62 / 22 (was 2.4.18) Parallel to x - 2 y = -5 containing (0,0) **** Give your equation for the requested line and explain how you obtained it.
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RESPONSE --> Slope is 1 y - y1= m(x -x1) y-0 = 1(x-0) y = 1x
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21:02:42 ** The equation x - 2y = -5 can be solved for y to give us y = 1/2 x + 5/2. A line parallel to this will therefore have slope 1/2. Point-slope form gives us y - 0 = 1/2 * (x - 0) or just y = 1/2 x. **
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RESPONSE --> Slope is 1/2 instead of 2/1.
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21:07:50 **** query 2.3.68 / 28 (was 2.4.24) Perpendicular to x - 2 y = -5 containing (0,4) **** Give your equation for the requested line and explain how you obtained it.
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RESPONSE --> x-2y = -5 (0, 4) slope 1/2 y-y1 = 1/2 (x-x1) y-4 = 1/2 ( x-0) y-4 = 1/2 x -0 y= 1/2x +4
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21:09:17 ** The equation x - 2y = -5 can be solved for y to give us y = 1/2 x + 5/2. A line perpendicular to this will therefore have slope -2/1 = -2. Point-slope form gives us y - 4 = -2 * (x - 0) or y = -2 x + 4. **
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RESPONSE --> Somehow I carried my 1/2 all through the equation.
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