#$&* course Mth 158 10/19/11 at 3:15 pm If your solution to stated problem does not match the given solution, you should self-critique per instructions at http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm.
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Given Solution: * * STUDENT SOLUTION: (-3,2) slope 4/3. Move 3 units in the x direction, 4 in the y direction to get ((-3+3), (2+4)), which simplifies to (0,6) (-3,2) slope 4/3 = -4/-3 so move -3 units in the x direction and -4 in the y direction to get ((-3-3), (2-4)) which simplifies to (-6,-2) From (0,6) with slope 4/3 we move 4 units in the y direction and 3 in the x direction to get ((0+3), (6+4)), which simplifies to (3,10). The three points I obtained are (-6,-2), (0,6), (3,10). * 2.3.40 / 36 (was 2.3.30). Line thru (-1,1) and (2,2) **** Give the equation of the line and explain how you found the equation. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: first I found the slope with the equation M = y2-y1/x2-x1 = 1/3 then I substitute 1/3 into the equation of a line y = mx + b and we substitute 0 in for b which = y = 1/3x + 0 the equation of the line is y-1/3x = 0 or y = 1/3x confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: * 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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: take the x-int and make a point (-4,0) and the y-int (0,4) then find the slope with the equation m = y2-y1/x2-x1 which = -1 then we substitute 4 in for b. the equation of the line is y = -1x + 4 or y + 1x = 4 confidence rating #$&*:1 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * STUDENT SOLUTION: The two points are (0, 4) and (4, 0). The slope is therefore m=rise / run = (4-0)/(0+4) = 1. The slope-intercept form is therefore y = m x + b = 1 x + 4, simplifying to y=x+4. STUDENT QUESTION I obtained -x + y = 4 or y = x + 4. I followed the example in the book which leaves 2 solutions (example problem 2.3.51) Did I do it correctly? INSTRUCTOR RESPONSE Both your solutions represent the same line, and both are correct. y = 1x + 4 means the same thing as y = x + 4; we rearrange this to -x + y = 4 (just subtract x from both sides). -x + y = 4 is a 'standard form' of the equation of this line. y = x + 4 is the 'slope-intercept' form of the equation. You don't need to know this, but still another 'standard form' is obtained by subtracting 4 from both sides of the equation -x + y = 4, giving us -x + y - 4 = 0. In this form we often want the coefficient of x to be positive, so we multiply both sides by -1 to get x - y + 4 = 0. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: * 2.3.76 / 56 (was 2.4.48). y = 2x + 1/2. **** What are the slope and the y-intercept of your line and how did you find them? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: the slope = 2 the y-int = 1/2 confidence rating #$&*:2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * 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.** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: * 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. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: first you put the equation in the form of the equation of the line. -2y = -1x - 5 y = 1/2x + 5/2 then we take the eqation y - y1 = m(x - x1) y - (0) = 1/2(x - (0)) y = 1/2x confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * The equation x - 2y = -5 can be solved for y to give us y = 1/2 x + 5/2. A line parallel to this will therefore have slope 1/2. Point-slope form gives us y - 0 = 1/2 * (x - 0) or just y = 1/2 x. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: * 2.3.68 / 28 (was 2.4.24) Perpendicular to x - 2 y = -5 containing (0,4) **** Give your equation for the requested line and explain how you obtained it. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: first you take the equation and put it in the equation of a line. x - 2y = -5 -2y = -x - 5 y = 1/2x + 5/2 y - (4) = -2(x - (0)) y = -2x + 4 confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: * * 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!