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Math Help - Find an equation of the line that satisfies the given conditions

  1. #1
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    Exclamation Find an equation of the line that satisfies the given conditions

    I have three "find the equation of the line" problems I cannot complete correctly.

    1.
    Through points (-1, -6) and (2, -4)

    2.
    Through the origin; parallel to the line 3x + 15y = 22

    --- on this one, I was able to find the slope. m= -3/15 but I got stuck after that.

    3.
    Through (-1, -2); perpendicular to the line 2x + 5y + 8 = 0

    ---I also found the slope for this one: m= 5/2, but again, I got stuck after.

    If anyone could help me with these, I would be grateful.

    Thanks!
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  2. #2
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    Quote Originally Posted by rckstr4lfe View Post
    I have three "find the equation of the line" problems I cannot complete correctly.

    1.
    Through points (-1, -6) and (2, -4)

    2.
    Through the origin; parallel to the line 3x + 15y = 22

    --- on this one, I was able to find the slope. m= -3/15 but I got stuck after that.

    3.
    Through (-1, -2); perpendicular to the line 2x + 5y + 8 = 0

    ---I also found the slope for this one: m= 5/2, but again, I got stuck after.

    If anyone could help me with these, I would be grateful.

    Thanks!
    to #1:

    If you have 2 points P_1(x_1, y_1) and P_2(x_2, y_2) then this is the equation of the line

    P_1P_2: \dfrac{y-y_1}{x-x_1} = \underbrace{\dfrac{y_2-y_1}{x_2-x_1}}_{slope\ of\ line}
    ----------------------------------------------------------------------------------------

    to #2:

    A parallel to 3x + 15y = 22 has the equation 3x + 15y = c . Plug in the coordinates of the origin to calculate the value of c.
    ----------------------------------------------------------------------------------------

    to #3:
    If you have a points P_1(x_1, y_1) and the slope of the line then this is the equation of the line

    P_1m: \dfrac{y-y_1}{x-x_1} = m~\implies~y-y_1=m(x-x_1)

    Plug in the values you already know.
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