25 Aug Clay This week’s discussion requires me to find the equations of
- Clay
- This week’s discussion requires me to find the equations of lines that are either parallel or perpendicular to the lines given. #19.
- My equation is y = -3x – 6; (-3,2)
- On the graph, this would be one point down on the x-axis origin and 5 points over
- y= mx+b
- 2=-3(-3)+b First, I will substitute the variables for x (-3) and y (2) into the equation.
- 2=-9+b
- 11=b Subtract 3 from each side to isolate the variable b or y–intercept
- The parallel equation is y = -3x+11
- 2=1/3 (-3) +b Giving a new slope of 1/3
- 2= -1/3 +b Substituting the variables for x (-3) and y (2) add 1/3 to both sides
- 2 +1/3 =b Add 1/3 to both sides to isolate the variable b.
- 2 1/3 = b
- The perpendicular equation is y= 1/3x + 2 1/3
- Reginald
- This week’s discussion requires me to find the equations of lines that are either parallel or perpendicular to the lines given. #12.
- My equation is y = -3x – 6; (-1, 5)
- On the graph, this would be one point down on the x-axis origin and 5 points over
- y= mx+b
- 5=-3(-1)+b First, I will substitute the variables for x (-1) and y (5) into the equation.
- 5= 3+b
- 2=b Subtract 3 from each side to isolate the variable b or y–intercept
- The parallel equation is y = -3x+2
- 5=1/3 (-1) +b Giving a new slope of 1/3
- 5= -1/3 +b Substituting the variables for x (-1) and y (5)
- add 1/3 to both sides
- 15/3 +1/3 =b Add 1/3 to both sides to isolate the variable b.
- 16/3 = b
- The perpendicular equation is y= 1/3x + 16/3
Respond to at least two of your classmates’ posts in at least a paragraph. Do you agree with how they used the vocabulary? Do their equations seem reasonable given what they started with?
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