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Thread: The line (-2,y) (-4,4) is perpendicular to the line (-1,-2) (4,-1). What is y?

  1. #1

    The line (-2,y) (-4,4) is perpendicular to the line (-1,-2) (4,-1). What is y?

    The line (-2,y) (-4,4) is perpendicular to the line (-1,-2) (4,-1). What is y?

    Here is how I have been attempting to work this out: y-4/-2-(-4) x -2-(-1)/-1-4 =-1 . From this I get: y-4/2 x -1/-5= -1.
    From this I get: 10 x y-4/2 x 10 x 1/5 = -1. From this I get: 5y-20 x 2 = -10. From this I get: 10y -40 = -10
    From this I get 10y = 30. An From this I conclude: y =3. But that is not what I need. I need a -6. But I can't seem to get that algebraically. So.....what am I doing wrong here? I spent an hour today trying to get this by algebraic means with the same result.

  2. #2
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    Quote Originally Posted by allegansveritatem View Post
    The line (-2,y) (-4,4) is perpendicular to the line (-1,-2) (4,-1). What is y?

    Here is how I have been attempting to work this out: y-4/-2-(-4) x -2-(-1)/-1-4 =-1 . From this I get: y-4/2 x -1/-5= -1.
    From this I get: 10 x y-4/2 x 10 x 1/5 = -1. I CAN'T FIGURE OUT WHAT IS GOING ON HERE. From this I get: 5y-20 x 2 = -10. From this I get: 10y -40 = -10
    From this I get 10y = 30. An From this I conclude: y =3. But that is not what I need. I need a -6. But I can't seem to get that algebraically. So.....what am I doing wrong here? I spent an hour today trying to get this by algebraic means with the same result.
    So the equation of the line formed by (-1, -2) and (4, -1) derives from

    [tex]\dfrac{y - (-\ 2)}{x - (-\ 1)} = \dfrac{-\ 2 - (-\ 1)}{-\ 1 - 4}.[/tex] Why?

    [tex]\dfrac{y - (-\ 2)}{x - (-\ 1)} = \dfrac{-\ 2 + 1}{-\ 1 - 4} \implies \dfrac{y + 2}{x + 1} = \dfrac{-\ 1}{-\ 5} = \dfrac{1}{5} \implies[/tex]

    [tex]5y + 10 = x + 1 \implies y = -\ \dfrac{9}{5} + \dfrac{1}{5} \cdot x.[/tex] Follow that?

    Let's check. If x is - 1,

    [tex]y = -\ \dfrac{9}{5} + \dfrac{1}{5} \cdot (-\ 1) = -\ \dfrac{9 + 1}{5} = -\ 2.[/tex] Checks.

    If x = 4,

    [tex]y = -\ \dfrac{9}{5} + \dfrac{1}{5} \cdot 4 = -\ \dfrac{9 - 4}{5} = -\ 1.[/tex] Checks.

    Now what is the slope of any line perpendicular to [tex]y = -\ \dfrac{9}{5} + \dfrac{1}{5} \cdot x?[/tex]

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    I don't quite follow your calculations and don't understand your notation. Are you familiar with vector operations?

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    Quote Originally Posted by allegansveritatem View Post
    The line (-2,y) (-4,4) is perpendicular to the line (-1,-2) (4,-1). What is y?

    Here is how I have been attempting to work this out: y-4/-2-(-4) x -2-(-1)/-1-4 =-1 . From this I get: y-4/2 x -1/-5= -1.
    From this I get: 10 x y-4/2 x 10 x 1/5 = -1. From this I get: 5y-20 x 2 = -10. From this I get: 10y -40 = -10
    From this I get 10y = 30. An From this I conclude: y =3. But that is not what I need. I need a -6. But I can't seem to get that algebraically. So.....what am I doing wrong here? I spent an hour today trying to get this by algebraic means with the same result.
    Let's clarify what you are doing, and then find your error (which is pretty simple) rather than find a different way to do it.

    You forgot to use parentheses to show the order of operations, which makes your work very hard to follow; also, you used "x" for the multiplication symbol, which is not a good idea in algebra where x is usually a variable. I'll use "*", which is standard in typing. So here is what I think you meant:

    (y-4)/(-2-(-4)) * (-2-(-1))/(-1-4) = -1 .

    From this I get: (y-4)/2 * -1/-5= -1.

    From this I get: 10 * (y-4)/2 * 10 * 1/5 = -1.

    The first two lines are correct; you are finding the slopes of the lines, and writing an equation that says their product is -1.

    The next line is wrong. You seem to be multiplying each of the factors on the left by 10, and not changing the right side! I suppose your idea is to use the LCD; but that is not needed for a product (only for a sum), and when you use it, you have to make sure you multiply the entire value on each side by the same thing.

    In effect, by multiplying twice on the left by 10, you have multiplied by 100, so you should be multiplying the right by 100 also.

    I would not do that; I would just multiply the fractions on the left together, and then multiply both sides by the new denominator.

  5. #5
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    Quote Originally Posted by allegansveritatem View Post
    The line (-2,y) (-4,4) is perpendicular to the line (-1,-2) (4,-1). What is y?

    Here is how I have been attempting to work this out: y-4/-2-(-4) x -2-(-1)/-1-4 =-1 . From this I get: y-4/2 x -1/-5= -1.
    From this I get: 10 x y-4/2 x 10 x 1/5 = -1. From this I get: 5y-20 x 2 = -10. From this I get: 10y -40 = -10
    From this I get 10y = 30. An From this I conclude: y =3. But that is not what I need. I need a -6. But I can't seem to get that algebraically. So.....what am I doing wrong here? I spent an hour today trying to get this by algebraic means with the same result.
    I usually don't bother doing this but I need to repeat (and you MUST see it!!) what Dr Peterson said: You multiplied the left side by 10 twice and did not multiply the right side by anything.

    Consider this: 2*6 =12 Easy and 100*12=1200
    Now 100*(2*6) = (10*10)*(2*6)= (10*2)*(10*6)=20*60=1200, that is (10*2)*(10*6)=100*(2*6). I multiplied both factors (2 and 6) by 10, just like you did.
    A mathematician is a blind man in a dark room looking for a black cat which isnít there. - Charles R. Darwin

  6. #6
    Quote Originally Posted by Dr.Peterson View Post
    Let's clarify what you are doing, and then find your error (which is pretty simple) rather than find a different way to do it.

    You forgot to use parentheses to show the order of operations, which makes your work very hard to follow; also, you used "x" for the multiplication symbol, which is not a good idea in algebra where x is usually a variable. I'll use "*", which is standard in typing. So here is what I think you meant:

    (y-4)/(-2-(-4)) * (-2-(-1))/(-1-4) = -1 .

    From this I get: (y-4)/2 * -1/-5= -1.

    From this I get: 10 * (y-4)/2 * 10 * 1/5 = -1.

    The first two lines are correct; you are finding the slopes of the lines, and writing an equation that says their product is -1.

    The next line is wrong. You seem to be multiplying each of the factors on the left by 10, and not changing the right side! I suppose your idea is to use the LCD; but that is not needed for a product (only for a sum), and when you use it, you have to make sure you multiply the entire value on each side by the same thing.

    In effect, by multiplying twice on the left by 10, you have multiplied by 100, so you should be multiplying the right by 100 also.

    I would not do that; I would just multiply the fractions on the left together, and then multiply both sides by the new denominator.
    I know what you are referring to here. In the exposition that I gave of my work I did neglect to show that I had taken that right side into account, but in the next line I show the fact that all terms have been multiplied by ten. The reason I used this method was to get rid of the fractions. Read through the rest of my work and you will see that the (-1) as been dealt with properly....but even so I do not get the right answer, which has to be (-6). I keep getting a 3. What is going on?

  7. #7
    Quote Originally Posted by Jomo View Post
    I usually don't bother doing this but I need to repeat (and you MUST see it!!) what Dr Peterson said: You multiplied the left side by 10 twice and did not multiply the right side by anything.

    Consider this: 2*6 =12 Easy and 100*12=1200
    Now 100*(2*6) = (10*10)*(2*6)= (10*2)*(10*6)=20*60=1200, that is (10*2)*(10*6)=100*(2*6). I multiplied both factors (2 and 6) by 10, just like you did.
    I just got done replying to Dr Perteson and now reading your reply I again see that you are saying I multiplied, in effect, by 100 on one side and by nothing on the other. But,as I said to Dr Peterson the fact that I wrote =-1 on the other side was a typo that the next line corrected. I don't understand what you mean by saying I multiplied by 100. How?

  8. #8
    Quote Originally Posted by allegansveritatem View Post
    I just got done replying to Dr Perteson and now reading your reply I again see that you are saying I multiplied, in effect, by 100 on one side and by nothing on the other. But,as I said to Dr Peterson the fact that I wrote =-1 on the other side was a typo that the next line corrected. I don't understand what you mean by saying I multiplied by 100. How?
    are you saying that I took one term to be two and multiplied each by 10? Is that it?

  9. #9
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    Quote Originally Posted by allegansveritatem View Post
    I know what you are referring to here. In the exposition that I gave of my work I did neglect to show that I had taken that right side into account, but in the next line I show the fact that all terms have been multiplied by ten. The reason I used this method was to get rid of the fractions. Read through the rest of my work and you will see that the (-1) as been dealt with properly....but even so I do not get the right answer, which has to be (-6). I keep getting a 3. What is going on?
    Okay, part of the problem was a typo that you then corrected, but not all.

    You said this:

    Quote Originally Posted by allegansveritatem View Post
    From this I get: y-4/2 x -1/-5= -1.
    From this I get: 10 x (y-4)/2 x 10 x 1/5 = -10. <--- Here you multiplied by 10 twice on the left
    From this I get: (5y-20) x 2 = -10. <--- Here you corrected the typo above.
    The trouble is that the two things on the left are factors, not terms. More below.

    Quote Originally Posted by allegansveritatem View Post
    I just got done replying to Dr Perteson and now reading your reply I again see that you are saying I multiplied, in effect, by 100 on one side and by nothing on the other. But,as I said to Dr Peterson the fact that I wrote =-1 on the other side was a typo that the next line corrected. I don't understand what you mean by saying I multiplied by 100. How?
    Quote Originally Posted by allegansveritatem View Post
    are you saying that I took one term to be two and multiplied each by 10? Is that it?
    Right, more or less.

    This is a common error; students get so used to distributing (multiplying each term) that they do it even when there is no addition. (Terms are things that are added.)

    When you multiply a sum of terms, you distribute:

    a(b + c) = ab + ac.

    But when you multiply a product, you just multiply once:

    a(bc) = abc, not abac.

    What you did was to multiply by 10 twice:

    (10a) * (10b)= 10*a*10*b = 10*10*a*b = 100ab.

    so you really multiplied by 100.

    Incidentally, this a reason to avoid writing multiplication symbols entirely; ab looks more like one term than a * b does, which is easily mistaken for a + b.

    Is it clear now?

  10. #10
    Quote Originally Posted by Dr.Peterson View Post
    Okay, part of the problem was a typo that you then corrected, but not all.

    You said this:



    The trouble is that the two things on the left are factors, not terms. More below.





    Right, more or less.

    This is a common error; students get so used to distributing (multiplying each term) that they do it even when there is no addition. (Terms are things that are added.)

    When you multiply a sum of terms, you distribute:

    a(b + c) = ab + ac.

    But when you multiply a product, you just multiply once:

    a(bc) = abc, not abac.

    What you did was to multiply by 10 twice:

    (10a) * (10b)= 10*a*10*b = 10*10*a*b = 100ab.

    so you really multiplied by 100.

    Incidentally, this a reason to avoid writing multiplication symbols entirely; ab looks more like one term than a * b does, which is easily mistaken for a + b.

    Is it clear now?
    yes, you are saying that I treated factors like addends and they are two very different animals. This is something pretty basic that I really wasn't aware of. It is very late and I don't have time to absorb this now, but tomorrow I will fool with it until it gets established. Thank you very much for pointing me in the right direction.

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