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Math Help - coordinate geometry.....

  1. #1
    Junior Member Bartimaeus's Avatar
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    Question coordinate geometry.....

    1. Point A lies on the line 3x+2y=24. A perpendicular is drawn from A to the x-axis, meeting at B
    a) if B=(a,0) find the coordinates of A in terms of a
    b) The triangle bounded by AB, 3x+2y=24 and the x-axis has an area of 27 square units. Find coordinates of A

    2.a) Solve equations y=3x^2 and y=12x-12 simulatneously
    b) explain why y=12x-12 is a tangent to parabola y=3x^2

    All working appreciated
    Thanks in advance!!!!!!!
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  2. #2
    Super Member wingless's Avatar
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    1.


    You can easily see that x coordinates of B and A are equal. Thus we know that A is (a,something). This is enough because all points on the line satisfies this equation: 3x+2y=24. We know x=a, solve the equation for y.



    2.
    y=3x^2
    y=12x-12

    3x^2=12x-12

    x^2-4x+4 = 0

    (x-2)^2=0

    x=2

    It means two graphs intersect only at x=2. There only one intersection. But, when a parabola and a line intersects at only one point, it's more like.. touching. Take a look at the graph and you'll easily see what I mean:



    So, the line is tangent to the parabola.
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  3. #3
    MHF Contributor Reckoner's Avatar
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    Quote Originally Posted by Bartimaeus View Post
    1. Point A lies on the line 3x+2y=24. A perpendicular is drawn from A to the x-axis, meeting at B
    a) if B=(a,0) find the coordinates of A in terms of a
    Check out this crude drawing:

    Code:
      \    / 
       \  /
       A\/
        /\
       /  \ 3x + 2y = 24
      /    \
    -o------\-----
    B=(a, 0) \
    We know the perpendicular must be of the form y = \frac23x + b (since the two slopes must multiply to -1). Since the x-intercept is at (a,\;0), we have

    0 = \frac23a + b\Rightarrow b = -\frac23a

    So the perpendicular line is y = \frac23(x - a). The intersection of these two lines will be A: find it.

    Quote Originally Posted by Bartimaeus View Post
    b) The triangle bounded by AB, 3x+2y=24 and the x-axis has an area of 27 square units. Find coordinates of A
    Here's a picture:

    Code:
           \
            o A
           /|\
          / | \
         /  |  \
        /  h|   \ 3x + 2y = 24
       /    |    \
      /     |     \ (8,0)
    -o-------------o--
    B       b       \
    You should have gotten \left(\frac{4a+72}{13},\;\frac{48-6a}{13}\right) for A. Thus we have

    \text{Area }=\frac12bh = \frac{24-3a}{13}\left(8 - a\right) = 27

    \Rightarrow \frac{3(8 - a)(8 - a)}{13} = 27\Rightarrow(8-a)^2 = 117.

    You should be able to take it from here.

    Quote Originally Posted by Bartimaeus View Post
    2.a) Solve equations y=3x^2 and y=12x-12 simulatneously
    b) explain why y=12x-12 is a tangent to parabola y=3x^2
    wingless has this covered.

    I realize that I may have misinterpreted question 1, but I'll leave my work as it is since I'm not sure.
    Last edited by Reckoner; June 12th 2008 at 01:15 AM.
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  4. #4
    Junior Member Bartimaeus's Avatar
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    Question

    My interpretation of the perpendicular part is that the perpendicular drawn is perpendicular to 3x+2y=24, ie, gradient of -(1/3), am I correct?
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  5. #5
    Super Member wingless's Avatar
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    Quote Originally Posted by Bartimaeus View Post
    My interpretation of the perpendicular part is that the perpendicular drawn is perpendicular to 3x+2y=24, ie, gradient of -(1/3), am I correct?
    Ah, I probably misunderstood that. See Reckoner's solution, he got it correct.
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  6. #6
    Junior Member Bartimaeus's Avatar
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    Question

    Quote Originally Posted by Reckoner View Post

    \text{Area }=\frac12bh = \frac{24-3a}{13}\left(8 - a\right) = 27

    \Rightarrow \frac{3(8 - a)(8 - a)}{13} = 27\Rightarrow(8-a)^2 = 117.
    okay, how did you go from \Rightarrow \frac{3(8 - a)(8 - a)}{13} = 27 to \Rightarrow(8-a)^2 = 117?
    where did the 24 go? and the - in front of the 3a from th previous line? and why (8-a)(8-a)?
    It doesn't quite add up to me...please explain further and add in any working you may have left out

    sorry for any inconvenience
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  7. #7
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    Quote Originally Posted by Bartimaeus View Post
    okay, how did you go from \Rightarrow \frac{3(8 - a)(8 - a)}{13} = 27 to \Rightarrow(8-a)^2 = 117?
    where did the 24 go? and the - in front of the 3a from th previous line? and why (8-a)(8-a)?
    It doesn't quite add up to me...please explain further and add in any working you may have left out

    sorry for any inconvenience

    24 - 3a = 3(8 - a)
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  8. #8
    MHF Contributor Reckoner's Avatar
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    Quote Originally Posted by Bartimaeus View Post
    okay, how did you go from \Rightarrow \frac{3(8 - a)(8 - a)}{13} = 27 to \Rightarrow(8-a)^2 = 117?
    where did the 24 go? and the - in front of the 3a from th previous line? and why (8-a)(8-a)?
    It doesn't quite add up to me...please explain further and add in any working you may have left out

    sorry for any inconvenience
    No problem. I'll start back at the beginning of that one. Did you get the same answer that I got for A, \left(\frac{4a+72}{13},\;\frac{48-6a}{13}\right)? Anyway, we have

    \text{\color{red}Area }=\frac12{\color{blue}b}{\color{magenta}h} (Formula for area of a triangle)

    \Rightarrow{\color{red}27} = \frac12{\color{blue}(8 - a)}{\color{magenta}\frac{48 - 6a}{13}} (Substitution)

    \Rightarrow\frac{(48 - 6a)(8 - a)}{26} = 27 (Simplifying)

    \Rightarrow\frac{6(8 - a)(8 - a)}{26} = 27 (Factoring: 48 - 6a = 6(8 - a))

    \Rightarrow\frac{3(8 - a)(8 - a)}{13} = 27 (Reducing fraction)

    \Rightarrow\frac{3(8 - a)^2}{13} = 27 (Rewriting with exponent)

    \Rightarrow(8 - a)^2 = \frac{27\cdot13}3 = 117 (Multiplying through by \frac{13}3)

    \Rightarrow8 - a = \pm\sqrt{117}\Rightarrow a = 8\pm\sqrt{117}

    Either solution should work (it depends on which side of the x-axis A is on; everything works out the same either way). Is that clearer?
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  9. #9
    Junior Member Bartimaeus's Avatar
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    Quote Originally Posted by Reckoner View Post
    b = -\frac23a[/tex]

    So the perpendicular line is y = \frac23(x - a). The intersection of these two lines will be A: find it.
    how do you find the y value? of A that is
    Last edited by Bartimaeus; June 19th 2008 at 01:07 AM. Reason: put in more detail
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  10. #10
    MHF Contributor Reckoner's Avatar
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    Quote Originally Posted by Bartimaeus View Post
    how do you find the y value? of A that is
    Do you understand how I found the equation of the perpendicular line? If so, we have two lines intersecting at A:

    3x + 2y = 24

    y = \frac23(x - a)

    All you need to do is find their intersection, which involves solving these two equations simultaneously. Rewriting the second equation produces

    3y = 2(x - a)\Rightarrow2x - 3y = 2a

    So now we have

    \left\{\begin{array}{rcrcrl}<br />
3x & + & 2y & = & 24 & \quad\text{\color{red}(1)}\\<br />
2x & - & 3y & = & 2a & \quad\text{\color{red}(2)}<br />
\end{array}\right.

    Multiply (1) by 2 and multiply (2) by 3:

    \left\{\begin{array}{rcrcrl}<br />
6x & + & 4y & = & 48 & \quad\text{\color{red}(3)}\\<br />
6x & - & 9y & = & 6a & \quad\text{\color{red}(4)}<br />
\end{array}\right.

    Now subtract (4) from (3):

    0x + 13y = 48 - 6a\Rightarrow y = \frac{48 - 6a}{13}

    This is our value for y. To get x, we back-substitute:

    3x + 2y = 24

    \Rightarrow3x + 2\left(\frac{48-6a}{13}\right) = 24

    \Rightarrow3x + \frac{12(8-a)}{13} = 24 (Factoring)

    \Rightarrow x = 8 - \frac{4(8 - a)}{13} (Dividing by 3 and subtracting the fraction)

    \Rightarrow x = \frac{104 - 4(8 - a)}{13} (Combining fractions)

    \Rightarrow x = \frac{4a + 72}{13} (Expanding and simplifying)

    Thus, the coordinates of A are \left(\frac{4a + 72}{13},\;\frac{48 - 6a}{13}\right), which agrees with the answer I gave earlier.
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