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Math Help - Find x coordinate of a point on a line

  1. #1
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    Find x coordinate of a point on a line

    Find k so that the line through (4, -1) and (k, 2) is
    a) parallel to 3y + 2x = 6
    b) perpendicular to 2y - 5x = 1

    How do I go about doing this? I think I have to get the given equation into slope-intercept form to get the slope but not sure how to get a coordinate of a point. Any ideas?
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  2. #2
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    Write each of the lines in the \displaystyle y = mx + c form so you can read off their gradients.

    Then evaluate the gradient of the line through the two points.

    Parallel lines have the same gradient, and the gradients of perpendicular lines multiply to give -1.
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  3. #3
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    I'm sorry. I'm not familiar with the term gradient. I know when changing the lines to y = mx + b form they come out to
    y = 2/3 x + 2 and y = -2/5 x + 1/2

    After I get this how do I find k from the coordinate (k, 2)?
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  4. #4
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    The gradient is the slope of the line, or the change in the dependent variable per unit change of the dependent variable (e.g. the number of kms per hour).
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  5. #5
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    Sooooo if I use the slope formula y2 - y1 / x2 - x1 where x1 or x2 = k it should equal 2/3 which is the slope of the first line? hmm but that wouldn't work considering if you take y2 - y1 no matter which y coordinate you use first it doesn't equal a 2.... Any hints?
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  6. #6
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    That doesn't matter, take for example the equation \displaystyle \frac{4}{25} = \frac{8}{x}, it can still be solved because \displaystyle \frac{4}{25} = \frac{8}{50}...
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  7. #7
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    so slope \displaystyle \frac{2}{3} = \displaystyle \frac{y_{2} - y_{1}}{k - x_{1}}<br />
which is \displaystyle \frac{2 - (-1)}{n} let n = k - x_{1}

    so

    \displaystyle \frac{2 - (-1)}{n} = \frac{3}{n}

    Slope \displaystyle \frac{2}{3} = \frac{3}{n} and when you solve for n you get \displaystyle \frac{3}{4.5}

    If 4.5 = k - x_{1}
    4.5 = k - 4
    4.5 + 4 = k
    8.5 = k

    Is this right or did I do too much thinking for nothing? lol.
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  8. #8
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    Looks good to me, if the gradient was \displaystyle \frac{2}{3}, except that the gradient is actually \displaystyle -\frac{2}{3}.

    But the process is essentially the same.
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  9. #9
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    So the concept is right but the calculations are wrong? I can deal with that. As long as I know how to work it I'm fine! Thanks for the help! it only took an hour and a half for me to do half of a problem with help lol.
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