Assignment 18

course Mth 158

018. `* 18*********************************************

Question: * 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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Your solution:

Move 3 units in x axis and 4 in the y axis = (0,6) do the same for -4 and -3 so move -3 in x axis and -4 in y axis to get (-6,-2)

(0+3), (6+4) = (3,10) so points are (-6,-2), (0,6), (3,10)

Confidence Assessment:

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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.

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Your solution:

y - y1 = m (x - x1)

y-1=1/3(x+1)

y = 1/3 x + 4/3

Confidence Assessment:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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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.

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Self-critique (if necessary):

Self-critique Rating:

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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?

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Your solution:

Y= mx + b so 1x+4 so y=x+4

Confidence Assessment:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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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.

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Self-critique (if necessary):

Self-critique Rating:

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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?

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Your solution:

y = 2 (0) + 1/2 or y = 1/2. So the y-intercept is (0, 1/2) so a slope of 2

Confidence Assessment:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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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.**

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Self-critique (if necessary):

Self-critique Rating:

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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.

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Your solution:

Solve for y giving y = 1/2 x + 5/2 with slope of ½. Point slope form gives y=1/2x

Confidence Assessment:

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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. **

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Self-critique (if necessary):

Self-critique Rating:

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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.

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Your solution:

Solve for y giving y=1/2x + 5/2 with slope ½. Line perpendicular = slope of -2 so y=-2+4

Confidence Assessment:

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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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. **

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