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Homework Help: Determining the equation of a line

  1. Apr 19, 2016 #1
    • Thread moved from the technical forums, so no Homework Help Template is shown
    Determine the equation of a line for the following:
    a) The line that has an x-intercept of 6 and that passes through the point (2, –2)
    b)The line that never intersects 5xy + 3 = 0, and which is five grid squares lower than that line

    Relevant equation:
    y= mx+b

    Attempt at a solution:
    for a, I know that the x intercept is +6, and the line has to pass through the points (2, -2). Does this mean i have to graph the line first, then solve for the slope of the line (m) and use any point on the line (x,y) to solve for b once i solve for the slope:

    After graphing the line i found the slope 2/4 or 1/2. I had to find b so i put the values i got in standard form:
    2=1/2(-2) + b
    I simplified both sides of the equation.
    Simplify: 2=b−1
    I then flip the equation.
    I added 1 to both sides.
    My answer:

    Is that the correct way to do it??

    for b, i solved for the slope intercept form of the line, then i graphed it, and counted five grid squares down. Then, i calculated the slope and y-intercept of the new line:

    5xy + 3 = 0
    5xy = -3
    y = -5x -3
    y = -5x -3 /-1
    y = 5x + 3

    after graphing this line, i counted 5 square grids down and got new line that has a y-intercept of -2, and the same slope as the original line (+5)
    so the equation of the line for part b is:
    y=5x -2

    is this correct???
  2. jcsd
  3. Apr 19, 2016 #2


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    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    Part b is correct.

    Part a is not. You plugged in x and y incorrectly.
  4. Apr 19, 2016 #3


    Staff: Mentor

    B looks right, both lines have the same slope which means they are parallel and so will never intersect.

    For A, you can check your solution as you have two points that must be on the line (2,-2) and the x-intercept point. Do you know the coordinates of the x-intercept?
  5. Apr 19, 2016 #4
    is this the answer for part a?
    -2 = 2/4(2) +b
    -2 = 1 + b
    -3 = b

    so the equation of the line for part a is:
    y= 1/2x -3
    is this correct??
  6. Apr 19, 2016 #5


    Staff: Mentor

    Take the two points mentioned earlier and plug them back into the equation. This is how you check you answer. So plug in the X intercept point which you should kno it's coordinates since it's an intercept point on the X axis. Similarly for the other point if either fail then you know your solution is wrong and so it's time to review your steps.

    We can't do your homework for you.
  7. Apr 19, 2016 #6
    so i plugged the x-intercept point into my equation, and heres what i got:

    0 = 1/2(6) -3
    0 = 3-3
    0 = 0
    so because both sides of the equation are balanced, im assuming my equation is correct???
  8. Apr 19, 2016 #7


    Staff: Mentor

    Ok and when you plug in 2,-2 what do you get?

    If that balances too then you have demonstrated that your solution is correct.
  9. Apr 19, 2016 #8
    -2 = 1/2(2) -3
    -2 = 1 - 3
    -2 = -2

    It does balance, so I'm guessing that my solution is correct?
  10. Apr 19, 2016 #9


    Staff: Mentor

    As a post analysis of your problem review your original steps to understand where you went wrong. Many times students will commit the same mistakes over and over losing valuable points on an exam.

    My brother would do something similar, he'd solve the problem but make a mistake in the check math and go back to "fix" his answer. Once he saw what he was doing he spent more time checking his check too.
  11. Apr 19, 2016 #10


    Staff: Mentor

    Yes I think you're right.
  12. Apr 19, 2016 #11
    Thanks for the advice and the help!
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