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Thread: Equations of Lines

  1. #1
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    Equations of Lines

    I have a misunderstanding that at the moment I can't get clear in my mind, so any help would be appreciated

    I am to find the slope of a line.

    I have two points (1,4) , (-6,5)

    m = y2 - y1 / x2 - x1

    m = 5 - 4 / -6 - 1

    m = - 1/7

    I am also asked to find the midpoint of the line.

    A(x1 , y1), B(x2 , y2)

    ( 1/2 (x1 + x2) , 1/2(y1 + y2))

    A = 1/2 (1 + (-6)) , B = 1/2 (4 + 5)

    A = - 5/2 and B = 9/2

    I am asked to find the equation of the bisector, this is where I get confused because I am not sure if the result below is to the original line and not the bisector?

    Equation of the line

    y - y1 = m (x - x1)

    y - 4 = - 7(x - 1)

    y - 4 = - 7x + 7

    y = - 7x + 11

    Now if I put the original "x" value into the equation y = - 7 (1) + 11, this yealds a result of 4, which is correct because it is the y intercept above.

    Now the problem.

    If the perpendicular line, which is at 90 degrees to the original line has an opposite gradient, i.e.

    perpendicular line is;

    -1 / - 1/7 = 7

    then using

    y - y1 = m (x - x1)

    we get

    y - 4 = 7(x - 1)

    y - 4 = 7x - 7

    y = 7x - 7 + 4

    y = 7x - 3

    If I now plug in the "x" value above we get

    y = 7(1) - 3

    y = 4

    There is absolutely nothing worse than fumbling around with scraps of paper trying and trying to get it right, thinking you have got it right, but then find you are suppose to use the midpoints AS fractions in your solutions, which is where I am now stuck?

    Any help appreciated
    Last edited by Probability; 04-14-2012 at 11:06 AM.

  2. #2
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    Quote Originally Posted by Probability View Post
    I have a misunderstanding that at the moment I can't get clear in my mind, so any help would be appreciated

    I am to find the slope of a line.

    I have two points (1,4) , (-6,5)

    m = y2 - y1 / x2 - x1

    m = 5 - 4 / -6 - 1

    m = - 1/7 <--- correct

    I am also asked to find the midpoint of the line.

    A(x1 , y1), B(x2 , y2)

    ( 1/2 (x1 + x2) , 1/2(y1 + y2))

    A = 1/2 (1 + (-6)) , B = 1/2 (4 + 5)

    A = - 5/2 and B = 9/2

    I am asked to find the equation of the bisector <--- there are an unlimited number of bisectors. You must be more clear here. I assume that you mean a perpendicular bisector(?)
    , this is where I get confused because I am not sure if the result below is to the original line and not the bisector?

    Equation of the line <--- do you mean the line AB(?)

    y - y1 = m (x - x1)

    y - 4 = - 7(x - 1) <--- if this should be AB then the slope is -1/7

    y - 4 = - 7x + 7

    y = - 7x + 11

    Now if I put the original "x" value into the equation y = - 7 (1) + 11, this yealds a result of 4, which is correct because it is the y intercept above.

    Now the problem.

    If the perpendicular line <--- or do you mean the perpendicular bisector(?)
    , which is at 90 degrees to the original line has an opposite gradient, i.e.

    perpendicular line is;

    -1 / - 1/7 = 7 <--- correct

    then using

    y - y1 = m (x - x1)

    we get

    y - 4 = 7(x - 1)

    y - 4 = 7x - 7

    y = 7x - 7 + 4

    y = 7x - 3

    If I now plug in the "x" value above we get

    y = 7(1) - 3

    y = 4

    There is absolutely nothing worse than fumbling around with scraps of paper trying and trying to get it right, thinking you have got it right, but then find you are suppose to use the midpoints AS fractions in your solutions, which is where I am now stuck?

    Any help appreciated
    If - and only if - you want to get the equation of the perpendicular bisector you must use the midpoint of AB. Your perpendicular line passes through point A which is an endpoint of the line segment.

    By the way: A sketch would be very helpful to check your arithmetic results.
    Attached Images Attached Images

  3. #3
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    Quote Originally Posted by pappus View Post
    If - and only if - you want to get the equation of the perpendicular bisector you must use the midpoint of AB. Your perpendicular line passes through point A which is an endpoint of the line segment.

    By the way: A sketch would be very helpful to check your arithmetic results.
    Thank you for your help, yes my notation still requires some more experience. Your graph is excellent I was confused with the bisector because I was unsure if the bisector extended in both directions as shown in your graph, which could now mean that my equation of the bisector I found here;

    y = 7x + 22

    could actually be correct according to your graph.

    The problem I have been facing is that if the equation is correct which I am not sure, then the coordinates should be found which they are not?

    y = 7(1) + 22

    = 29

    the Y coordinate is 4

    This is why I am getting confused.

  4. #4
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    Quote Originally Posted by Probability View Post
    I have a misunderstanding that at the moment I can't get clear in my mind, so any help would be appreciated

    I am to find the slope of a line.

    I have two points (1,4) , (-6,5)

    m = y2 - y1 / x2 - x1

    m = 5 - 4 / -6 - 1

    m = - 1/7

    I am also asked to find the midpoint of the line.

    A(x1 , y1), B(x2 , y2)

    ( 1/2 (x1 + x2) , 1/2(y1 + y2))

    A = 1/2 (1 + (-6)) , B = 1/2 (4 + 5)

    A = - 5/2 and B = 9/2 The coordinates of the MIDPOINT are (-5/2, 9/2). I am not sure how you are using the letters A and B here.....

    I am asked to find the equation of the bisector, this is where I get confused because I am not sure if the result below is to the original line and not the bisector?

    Equation of the line

    y - y1 = m (x - x1)

    y - 4 = - 7(x - 1)

    y - 4 = - 7x + 7

    y = - 7x + 11

    Now if I put the original "x" value into the equation y = - 7 (1) + 11, this yealds a result of 4, which is correct because it is the y intercept above.

    Now the problem.

    If the perpendicular line, which is at 90 degrees to the original line has an opposite gradient, i.e.

    perpendicular line is;

    -1 / - 1/7 = 7

    then using

    y - y1 = m (x - x1)
    The perpendicular bisector of a segment must contain the MIDPOINT of the segment....which is (-5/2, 9/2)
    So, (x1, y1) is (-5/2, 9/2)


    we get

    y - 4 = 7(x - 1) No.....x1 = -5/2 and y1 = 9/2. So, you SHOULD have
    y - (9/2) = 7[x - (-5/2)]


    The rest of your work here needs to be adjusted accordingly.

    y - 4 = 7x - 7

    y = 7x - 7 + 4

    y = 7x - 3

    If I now plug in the "x" value above we get

    y = 7(1) - 3

    y = 4

    There is absolutely nothing worse than fumbling around with scraps of paper trying and trying to get it right, thinking you have got it right, but then find you are suppose to use the midpoints AS fractions in your solutions, which is where I am now stuck?

    Any help appreciated
    See my comments in red above.

    (x1, y1) as used in the point-slope formula is the point you know the line goes through. The perpendicular bisector of a segment must go through the midpoint of the segment...(-5/2, 9/2) is the midpoint of the segment you're working with.

    Didn't I a answer this question for you earlier, using the "slope-intercept" form for the equation of a line because that's what you "knew about"?
    Last edited by Mrspi; 04-15-2012 at 08:39 PM.

  5. #5
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    Quote Originally Posted by Mrspi View Post
    See my comments in red above.

    (x1, y1) as used in the point-slope formula is the point you know the line goes through. The perpendicular bisector of a segment must go through the midpoint of the segment...(-5/2, 9/2) is the midpoint of the segment you're working with.

    Didn't I a answer this question for you earlier, using the "slope-intercept" form for the equation of a line because that's what you "knew about"?



    Thank you for pointing the above out, I did realise this some time afterwards when I looked back at the question and the question prompted me to look at the midpoint rule, then I wondered why I had not included the fractions I wrote


    I have now sketched the graph and also continued on and now included a second perpendicular bisector, where more coordinates have been added. The graph you produced has A (1,4) and B (-6, 5). I have now included C (-7, 4). Between BC I have now included the next perpendicular bisector and it's gradient. Because BC is pointing downwards to the left in the second quadrant I am thinking the gradient of this perpendicular bisector is 1 rather than -1, or am I getting this the wrong way round?

    Using these two perpendicular bisectors I am now trying to find the centre of the circle using my previously calculated two equations, which were

    y = 7x + 44/2

    y = -1x + 22/2 (IF) this gradient is correct?

    I am however unsure about these two equations as the graph I have drawn does not concur with these equations as saying they represent the centre of the circle?

    So I am still learning

    Thanks for your help, if you could give any pointers I would be much appreciative of your input.

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