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Math Help - Slope zero??

  1. #1
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    Slope zero??

    Line l contains the points (5\sqrt{2}, -6) and (\sqrt{50}, 2). Give the slope of any line perpendicular to l.

    Ok. Perpendicular means that the product of their slopes must be equal to -1 as they are negative reciprocals.

    So:

    Slope = \frac{rise}{run} = \frac{y_1-y_2}{x_1-x_2} = \frac{2-(-6)}{\sqrt{50}-5\sqrt{2}}


    Slope = \frac{2+6}{(\sqrt{25})(\sqrt{2})-5\sqrt{2}}


    Slope = \frac{8}{5\sqrt{2}-5\sqrt{2}}


    Slope = \frac{8}{0}


    But then Division by zero does not count. The book says the answer should be 0. Did I do it right?. All comments are welcomed
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  2. #2
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    Re: Slope zero??

    You are aware that \sqrt{50}= \sqrt{25(2)}= 5\sqrt{2} are you not? So the two given points are (5\sqrt{2}, -6) and (5\sqrt{2}, 2). Since the x coordinates are the same, the line is vertical. Any line perpendicular to it is horizontal and so has 0 slope.

    The fact that a line "does not have a slope" does not mean a line perpendicular to is does not. Again, the fact that you were unable to calculate a slope for this line means it is vertical and so a line perpendicular to it is horizontal.

    (I would not say "division by 0 does not count". I would say "we cannot divide by 0 so this line has NO slope- which means it is vertical". And please use [ tex ] and [ /tex ] tags.)
    Last edited by HallsofIvy; July 10th 2014 at 05:42 AM.
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    Re: Slope zero??

    Quote Originally Posted by Alienis Back View Post
    Line l contains the points (5\sqrt{2}, -6) and (\sqrt{50}, 2). Give the slope of any line perpendicular to l.

    Ok. Perpendicular means that the product of their slopes must be equal to -1 as they are negative reciprocals.

    So:

    Slope = \frac{rise}{run} = \frac{y_1-y_2}{x_1-x_2} = \frac{2-(-6)}{\sqrt{50}-5\sqrt{2}}


    Slope = \frac{2+6}{(\sqrt{25})(\sqrt{2})-5\sqrt{2}}
    You have an error in this line. You never factored a sqrt{25} out of the second term. You will find, though, that your final answer is correct: You made the same mistake twice and it balanced out.

    So, your work is essentially correct. If the line has a "slope" of (something)/0 then what direction is the line in? Hint: rise/run for this line is what?

    -Dan

    (Ahhhh! He beat me!)
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    Re: Slope zero??

    I couldn't understand the question,Please elaborate

    Quote Originally Posted by Alienis Back View Post
    Line l contains the points (5\sqrt{2}, -6) and (\sqrt{50}, 2). Give the slope of any line perpendicular to l.

    Ok. Perpendicular means that the product of their slopes must be equal to -1 as they are negative reciprocals.

    So:

    Slope = \frac{rise}{run} = \frac{y_1-y_2}{x_1-x_2} = \frac{2-(-6)}{\sqrt{50}-5\sqrt{2}}


    Slope = \frac{2+6}{(\sqrt{25})(\sqrt{2})-5\sqrt{2}}


    Slope = \frac{8}{5\sqrt{2}-5\sqrt{2}}


    Slope = \frac{8}{0}


    But then Division by zero does not count. The book says the answer should be 0. Did I do it right?. All comments are welcomed
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    Re: Slope zero??

     Slope = \frac{rise}{run} = \frac{y_1-y_2}{x_1-x_2} = \frac{2-(-6)}{\sqrt{50}-5\sqrt{2}}

     Slope = \frac{2+6}{(\sqrt{25})(\sqrt{2})-5\sqrt{2}}


     Slope = \frac{8}{5\sqrt{2}-5\sqrt{2}}


     Slope = \frac{8}{0}
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  6. #6
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    Re: Slope zero??

    burgess, Ive been given a line  l , which crosses the follolowing two points:  (5\sqrt{2}, -6) and (\sqrt{50}, 2) . Then I have to find a line (another line) which is perpendicular to  l . You have to know that parallel lines have the same slope, and perpendicular lines have slop  -1 .

    For instance, if you draw a line with slope \frac{-4}{3}, then you will find that ANY line with slope \frac{3}{4} is going to be perpendicular to the first one, because they are negative reciprocals. Their product is -1.
    Last edited by Alienis Back; July 10th 2014 at 09:55 AM.
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  7. #7
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    Re: Slope zero??

    Did you not read what others have written here? The slope is NOT \frac{8}{0} because we cannot divide by 0. The slope does not exist which tells us that this is a vertical line. A line perpendicular to it is horizontal and so has slope 0.

    (Exactly what definition of "slope" are you using and what properties of slope have you learned? Slope can be defined as "tangent of the angle the line makes with the x-axis pr other horizontal line". Or if you have some other definition, you can show that as a property. Here, the line is vertical so it makes angle 90 degrees with the x-axis. tan(90) does not exist. A line perpendicular to that line is horizontal and makes angle 0 degrees with the x-axis. tan(0)= 0.)
    Last edited by HallsofIvy; July 10th 2014 at 09:59 AM.
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  8. #8
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    Re: Slope zero??

    Ok. I did read the post, and I know the expression \frac{8}{0} is undefined... I mean, there is no real number to associate with it. In this case the line is vertical, parallel to y axis, has NO slope. I also know ANY PERPENDICULAR will be parallel to the x axis and ALL horizontal lines have 0. Now, when I tryied to proof that mathematically and got \frac{8}{0} for an answer... I wasn't sure so I posted here. All I wanted to know, was whether my calculations were right or not...
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  9. #9
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    Re: Slope zero??

    To "solve" this problem, we need a "better definition" of perpendicular, one that lets us check "vertical lines", too. So let's stop thinking of lines as functions (because vertical lines AREN'T functions) and think of them THIS way:

    A LINE is a set in the plane:

    $L = \{(x,y) \in \Bbb R^2: Ax + By + C = 0\}$ for some real numbers $A,B,C$.

    Note that if $f(x) = mx + b$, and we look at the graph of $f$, which is the set of points $(x,f(x))$, this fits our "new definition", we can simply take:

    $A = m, B = -1, C = b$.

    For two lines. say:

    $L_1: A_1x + B_1y + C_1 = 0$ and

    $L_2: A_2x + B_2y + C_2 = 0$, we say $L_1$ and $L_2$ are perpendicular if: $A_1A_2 + B_1B_2 = 0$.

    Let's verify that this works with our "old" definition of perpendicular:

    Suppose we have:

    $f(x) = mx + b$

    $g(x) = \dfrac{-1}{m}x + b'$. Writing these two lines in our "new form", we have:

    $L_1: mx - y + b = 0$

    $L_2: \dfrac{-1}{m}x - y + b' = 0$. Computing $A_1A_2 + B_1B_2 = m\dfrac{-1}{m} + (-1)(-1) = -1 + 1 = 0$, we see all perpendicular lines under our OLD definition are STILL perpendicular under our NEW definition.

    What have we gained? Well, with our "new" definition, a horizontal line is one for which $A = 0$, and a vertical line is one for which $B = 0$. Checking to see if these are perpendicular, we have:

    $A_1A_2 + B_1B_2 = (0)(A_1) + (B_1)(0) = 0 + 0 = 0$. We have avoided the "divide by 0" problem. This is better.

    **********

    So now, given that $(5\sqrt{2},-6)$ and $(\sqrt{50},2)$ both lie on $L_1$, we set out to find $A_1,B_1$ and $C_1$.

    First, we see that:

    $A_1(5\sqrt{2}) + B_1(-6) + C_1 = A_1(\sqrt{50}) + B_1(2) + C_1$ (these are both 0). We can subtract $C_1$ from both sides, so it is irrelevant, here (we can find it later, if we need it).

    $A_1(5\sqrt{2} - \sqrt{50}) = 8B_1$ (simple arithmetic, nothing fancy). Note that $\sqrt{50} = \sqrt{25}\sqrt{2} = 5\sqrt{2}$, so we have:

    $A_1(0) = 8B_1$

    $0 = 8B_1$

    $0 = B_1$. We have a vertical line. We haven't found $A_1$, you can try, if you like. All we need to know is it isn't 0 (verify this!).

    Perhaps horizontal lines aren't the only ones perpendicular (under our new definition) to vertical ones. We should investigate this.

    According to our new definition, if $l :A_2x + B_2y + C_2 = 0$, we need to have $A_1A_2 + B_1B_2 = 0$. Since $A_1 \neq 0$, and $B_1 = 0$, this becomes:

    $A_1A_2 + 0 = 0$

    $A_2 = \dfrac{0}{A_1} = 0$, which is a horizontal line.

    I would regard all the previous answers with extreme suspicion, as they have asserted "horizontal lines are perpendicular to vertical lines" without defining these terms.
    Last edited by Deveno; July 10th 2014 at 02:14 PM.
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