Results 1 to 8 of 8

Math Help - Find the equation of the line

  1. #1
    Senior Member
    Joined
    Jul 2008
    Posts
    313

    Find the equation of the line

    I am presented with coordinates (5,3) and (-7, 7). These coordinates as I understand are;

    x = 5
    y = 3

    and

    x = -7
    y = 7

    I have produced a graph, see below.

    I am asked to find the equation of the line, but to do this I require to calculate the gradient of the line, which I have done as follows;

    Gradient = y2 - y1 = 7 - 3 = 4
    ...............x2 - x1.... -7 - 5 -12

    Gradient = - 0.33.

    I must be misunderstanding something in the presenting of the values because even I know that the gradient of that graph is more than -0.33?

    Even if I turn that data above around, i.e. 3 - 7 = - 4 = - 0.33
    .................................................. .....5 - (-7) 12

    I am clearly missing something here?

    The graph won't load up sorry?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,820
    Thanks
    1711
    Awards
    1

    Re: Find the equation of the line

    The slope is not -.33.
    It is \frac{-1}{3}. Always use fraction if possible.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member
    Joined
    Jul 2008
    Posts
    313

    Re: Find the equation of the line

    I will have to do some more research on this subject, as a fraction the 8/3 = 2
    .................................................. .................................................. ...2/3

    as a decimal this equates to 2.66 recurring.

    My coordinates of (5,3)(-7,7) don't reflect the answer found?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Apr 2005
    Posts
    16,038
    Thanks
    1675

    Re: Find the equation of the line

    Another way of doing this: Any (non-vertical) line can be written in the form y= ax+ b. Setting x= 5, y= 3, we have the equation 3= 5a+ b. Setting x= -7, y= 7, we have the equation 7= -7a+ b. Subtracting the second equation from the first eliminates b: 3- 7= 5a- (-7a). -4= (5+ 7)a, -4= 12a, a= -4/12= -1/3 (which is an exact answer where "a= -.33" is only approximate).

    Now, my question is, why do you say "even I know that the gradient of that graph is more than -0.33"? The line runs from (-7, 7) to the right down to (5, 3). With a "run" of 12 there has been a "rise' of -4. The line goes down 1/3 the distance it goes to the right. That surely looks like "-1/3" to me!
    Follow Math Help Forum on Facebook and Google+

  5. #5
    -1
    e^(i*pi)'s Avatar
    Joined
    Feb 2009
    From
    West Midlands, England
    Posts
    3,053
    Thanks
    1

    Re: Find the equation of the line

    What method have you learned for finding a straight line?

    I prefer the one which says y - y_1 = m(x-x_1) where (x_1,y_1) is a given coordinate and m is the gradient.

    To find the gradient use the equation m = \dfrac{y_2 - y_1}{x_2 - x_1} where (x_1,y_1) \text{  and  } (x_2,y_2) are two points on the line. As long as you are consistent with which point is 1 and which is 2 the order does not affect the answer which you found out in the OP.


    In this question we have the points (5,3) and (-7,7). Using the equation above [(-7,7) is (x_1,y_1) here] we work out the gradient \dfrac{3-7}{5-(-7)} = \dfrac{-4}{12} = -\dfrac{1}{3}.

    Spoiler:
    It doesn't matter because if we look at what happens when we multiply top and bottom by -1 (which we know doesn't change the value):

    \dfrac{-(y_2-y_1)}{-(x_2-x_1)} = \dfrac{-y_2 + y_1}{-x_2 + x_1} = \dfrac{y_1-y_2}{x_1-x_2}

    You don't need to know why it is so for this example, but as long as you're consistent you'll get the correct gradient


    Now we know that m = -\dfrac{1}{3} we can sub it in to the equation of a line using either of our coordinates. I shall choose (5,3) as it contains no negative numbers

    y - 3 = -\dfrac{1}{3}(x - 5) and upon adding three to both sides: y = 3 - \dfrac{1}{3}(x-5)

    it is up to you to arrange that into the form y=mx+c


    edit: perhaps Wolfram will be of use as a visual aid


    btw: if you don't use latex can you just use plain text with brackets please? Your working is almost impossible to read. (y2-y1)/(x2-x1) = m is much easier!
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Senior Member
    Joined
    Jul 2008
    Posts
    313

    Re: Find the equation of the line

    Quote Originally Posted by e^(i*pi) View Post
    What method have you learned for finding a straight line?

    I prefer the one which says y - y_1 = m(x-x_1) where (x_1,y_1) is a given coordinate and m is the gradient.

    To find the gradient use the equation m = \dfrac{y_2 - y_1}{x_2 - x_1} where (x_1,y_1) \text{ and } (x_2,y_2) are two points on the line. As long as you are consistent with which point is 1 and which is 2 the order does not affect the answer which you found out in the OP.


    In this question we have the points (5,3) and (-7,7). Using the equation above [(-7,7) is (x_1,y_1) here] we work out the gradient \dfrac{3-7}{5-(-7)} = \dfrac{-4}{12} = -\dfrac{1}{3}.

    Spoiler:
    It doesn't matter because if we look at what happens when we multiply top and bottom by -1 (which we know doesn't change the value):

    \dfrac{-(y_2-y_1)}{-(x_2-x_1)} = \dfrac{-y_2 + y_1}{-x_2 + x_1} = \dfrac{y_1-y_2}{x_1-x_2}

    You don't need to know why it is so for this example, but as long as you're consistent you'll get the correct gradient


    Now we know that m = -\dfrac{1}{3} we can sub it in to the equation of a line using either of our coordinates. I shall choose (5,3) as it contains no negative numbers

    y - 3 = -\dfrac{1}{3}(x - 5) and upon adding three to both sides: y = 3 - \dfrac{1}{3}(x-5)

    it is up to you to arrange that into the form y=mx+c


    edit: perhaps Wolfram will be of use as a visual aid


    btw: if you don't use latex can you just use plain text with brackets please? Your working is almost impossible to read. (y2-y1)/(x2-x1) = m is much easier!
    Thank you for your help on this thread.

    First in reply to another members thread, the reason I thought the gradient was wrong at -0.33 was because I compared that result to the graph that a produced and what looked back at me was in appearance significantly different, what I didn't think to do until it was pointed out was to use fractions, therefore;

    -0.33 = -1/3

    In relation to the work I did previously and the help received, which is greatly appreciated, however, the above technique I can't follow it the way it has been presented, my limited experience does not permit me now to understand what value has been chosen for x, as in my coordinates 5 was the x value?

    Using y = mx + c

    I tried the following.

    y = - 1/3 x -7 + 16/3 = 7/3 + 16/3 = 23/3 = 7.7

    This type of answer has been my problem all along, my coordinates are (5, 3) and (-7, 7)

    So getting answers to 1 d.p is throwing me off and I can't see where I am getting it wrong?

    So I keep asking myself am I getting something wrong or is there something wrong with the question being asked?

    Thanks

    David
    Follow Math Help Forum on Facebook and Google+

  7. #7
    -1
    e^(i*pi)'s Avatar
    Joined
    Feb 2009
    From
    West Midlands, England
    Posts
    3,053
    Thanks
    1

    Re: Find the equation of the line

    Quote Originally Posted by David Green View Post

    Using y = mx + c

    I tried the following.

    y = - 1/3 x -7 + 16/3 = 7/3 + 16/3 = 23/3 = 7.7

    This type of answer has been my problem all along, my coordinates are (5, 3) and (-7, 7)

    So getting answers to 1 d.p is throwing me off and I can't see where I am getting it wrong?

    So I keep asking myself am I getting something wrong or is there something wrong with the question being asked?

    Thanks

    David
    Where has 16/3 come from?

    As you said you have y=mx+c and it appears that you're using the co-ordinate (-7,7) which is fine so if sub in x=-7 and y=7 then you find c.

    7 = -\dfrac{1}{3}\times -7 + C = 7/3 + C

    C = 7 - \dfrac{7}{3} = \dfrac{21}{3} - \dfrac{7}{3} = \dfrac{14}{3}

    Thus your equation is y = -\dfrac{1}{3}x + \dfrac{14}{3}. You can test this by plugging in your other coordinate and seeing if the LHS=RHS
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Senior Member
    Joined
    Jul 2008
    Posts
    313

    Re: Find the equation of the line

    Quote Originally Posted by e^(i*pi) View Post
    Where has 16/3 come from?

    As you said you have y=mx+c and it appears that you're using the co-ordinate (-7,7) which is fine so if sub in x=-7 and y=7 then you find c.

    7 = -\dfrac{1}{3}\times -7 + C = 7/3 + C

    C = 7 - \dfrac{7}{3} = \dfrac{21}{3} - \dfrac{7}{3} = \dfrac{14}{3}

    Thus your equation is y = -\dfrac{1}{3}x + \dfrac{14}{3}. You can test this by plugging in your other coordinate and seeing if the LHS=RHS
    Thank you, it is a clear lack of understanding on my part.

    I knew how to calculate the gradient, but was confused with the result obtained, that was throwing me out first.

    Once that problem was solved I then hit the next hurdle which I thought I understood, but obviously not?

    I then started to get confused with the fractions because I had tried to find the solution in so many different ways I completely confused myself.

    In the workbook using y = mx + c, I failed to notice that "x" disapeared when the example was being worked out, therefore when I was trying to work out a solution I was including a value for "x", which was giving me incorrect results.

    This was part of the confusion in the book;

    y = mx + c

    gradient = 3/2

    y = 3/2(x) + c

    y = 3/2 x 1 + c

    Look at the co-ordinates they used in the example (1,2)

    So I started to use the same idea, however they did not include the (x 1)?

    The blind leading the blind.

    I know understand the method you have explained.

    Thank you

    David
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 8
    Last Post: March 22nd 2011, 04:57 PM
  2. Replies: 1
    Last Post: March 18th 2011, 10:36 AM
  3. Replies: 9
    Last Post: July 31st 2009, 05:39 AM
  4. Replies: 11
    Last Post: June 2nd 2009, 06:08 PM
  5. Replies: 5
    Last Post: October 13th 2008, 10:16 AM

Search Tags


/mathhelpforum @mathhelpforum