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Thread: the equation of a straight line

  1. Oct 13th 2008, 11:05 AM #1
    coyoteflare
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    the equation of a straight line

    the equation of stralight line is 3x+2y=11 find the equation of the parrallel line passing through the point 2,5

    i got y=2/3x +7 1/3

    or is it y=2/3x -4.5
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  2. Oct 13th 2008, 11:29 AM #2
    masters
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    Quote Originally Posted by coyoteflare View Post
    the equation of stralight line is 3x+2y=11 find the equation of the parrallel line passing through the point 2,5

    i got y=2/3x +7 1/3

    or is it y=2/3x -4.5
    3x+2y=11

    2y=-3x+11

    y=-\frac{3}{2}x+\frac{11}{2}

    Slope(m) = -\frac{3}{2}

    Use slope-intercept form of the linear equation: y=mx+b and (2, 5) and solve for b.

    y=mx+b

    5=-\frac{3}{2}(2)+b

    5=-3+b

    8=b

    Substituting back into y=mx+b, we have:

    y=-\frac{3}{2}x+8
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  3. Oct 13th 2008, 11:29 AM #3
    earboth
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    Quote Originally Posted by coyoteflare View Post
    the equation of stralight line is 3x+2y=11 find the equation of the parrallel line passing through the point 2,5

    i got y=2/3x +7 1/3

    or is it y=2/3x -4.5
    The LHS of the equation of the parallel must be 3x + 2y too. To calculate the RHS you only have to plug in the coordinates of the given point. Thus the equation of the parallel is:

    \boxed{3x+2y=16}
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  4. Oct 13th 2008, 11:35 AM #4
    coyoteflare
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    i feel so stupid sorry i meant..

    Quote Originally Posted by masters View Post
    3x+2y=11

    2y=-3x+11

    y=-\frac{3}{2}x+\frac{11}{2}

    Slope(m) = -\frac{3}{2}

    Use slope-intercept form of the linear equation: y=mx+b and (2, 5) and solve for b.

    y=mx+b

    5=-\frac{3}{2}(2)+b

    5=-3+b

    8=b

    Substituting back into y=mx+b, we have:

    y=-\frac{3}{2}x+8
    ..perpendicular.. sorry not parrallel...

    getting confused cus im trying to do binomials too
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  5. Oct 13th 2008, 11:56 AM #5
    masters
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    Quote Originally Posted by coyoteflare View Post
    ..perpendicular.. sorry not parrallel...

    getting confused cus im trying to do binomials too
    In that case, just change the slope to \frac{2}{3} which is the negative reciprocal of -\frac{3}{2} and then proceed the same way.

    5=\frac{2}{3}(2)+b

    5=\frac{4}{3}+b

    \frac{11}{3}=b

    The equation of the perpendicular is y=\frac{2}{3}x+\frac{11}{3}
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