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Math Help - equation of straight line

  1. #1
    Newbie fresh_'s Avatar
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    Exclamation equation of straight line

    hey,
    write down the equation of the straight line that:
    a) passes through (0,-10) and is parrallel to the line with equation y=6x+3
    b) passes through the point (0,-1) and is perpendicular to the line with equation 3x-2y+5=0


    thanks
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  2. #2
    Moo
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    Hello,

    Quote Originally Posted by fresh_ View Post
    hey,
    write down the equation of the straight line that:
    a) passes through (0,-10) and is parrallel to the line with equation y=6x+3
    If two lines are parallel, then the slopes are equal.

    The slope of y=6x+3 is 6.
    Therefore, the equation of the line we want to find is y=6x+a, with a to determine.
    Because it passes through (0,-10), we have -10=6*0+a. So a=-10.

    Thus the equatio of the line is y=6x-10

    b) passes through the point (0,-1) and is perpendicular to the line with equation 3x-2y+5=0
    y=ax+b

    It passes through the point 0,-1. Therefore, -1=a*0+b --> b=-1

    ax-y-1=0.

    A normal vector to it is (a,-1).

    A normal vector to 3x-2y+5=0 is (3,-2).

    Because the two lines are perpendicular, their normal vector are perpendicular too. In particular, the scalar product is 0.

    0=(a,-1) \dot (3,-2)=3a+2 \implies a=-\frac 23

    You have the equation of the line now;..

    i hope i didn't make any mistake...
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  3. #3
    Newbie fresh_'s Avatar
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    Quote Originally Posted by Moo View Post
    Hello,


    If two lines are parallel, then the slopes are equal.

    The slope of y=6x+3 is 6.
    Therefore, the equation of the line we want to find is y=6x+a, with a to determine.
    Because it passes through (0,-10), we have -10=6*0+a. So a=-10.

    Thus the equatio of the line is y=6x-10


    y=ax+b

    It passes through the point 0,-1. Therefore, -1=a*0+b --> b=-1

    ax-y-1=0.

    A normal vector to it is (a,-1).

    A normal vector to 3x-2y+5=0 is (3,-2).

    Because the two lines are perpendicular, their normal vector are perpendicular too. In particular, the scalar product is 0.

    0=(a,-1) \dot (3,-2)=3a+2 \implies a=-\frac 23

    You have the equation of the line now;..

    i hope i didn't make any mistake...
    so equation of line is y= - 2/3x + -1

    is their any way to arrange the equation to 3y + 2x =-3 ?
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  4. #4
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    @Moo: You're mistake-free.
    @fresh_: Yes, of course.

    Here I listed the steps for you:
    Given: y= -\frac{2x}{3} - 1

    Move 1 to the other side: y + 1 = -\frac{2x}{3}

    Cross multiply: 3y + 3 = -2x

    Switch sides with 2x and 3: 3y + 2x = -3

    OR you can simply do this:
    \left(y + \frac{2x}{3} =  - 1\right) \cdot 3

    3y + 2x = -3
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