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Math Help - Some problems

  1. #1
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    Some problems

    Hi I have problems in some questions.
    A) My answers to these questions do not match the book answers:-

    1) The three vertices of a paralleologram are (3,4), (3,8) and (9,8). Find the fourth vertex.[Book Answer: (9,4)](Solved)

    2) The area of a triangle is 5. Two of its vertices are (2,1) and (3,-2). The third vertex lies on y=x+3. Find the third vertex.[Book Answer: (7/2,13/2) and (-3/2,3/2). How am I supposed to get the second answer](Solved)

    3) The point A divides the join of P(-5,1) and Q(3,5) in the ratio k:1. Find the values of k for which the area of triangle ABC where B is (1,5) and C(7,-2) is equal to 2 units.[Book Answer: k=7 or 31/9]



    B) Can't even guess what to do:-

    1) Find the locus of the mid-point of the portion of the line x cos\alpha + y sin\alpha = p which is intercepted between the axes.[Book Answer: p^2(x^2+y^2)=4x^2y^2]

    2) Find the equation of the bisector of angle A of the triangle whose vertices are A(4,3), B(0,0) and C(2,3).[Book Answer: x-3y+5=0](Solved)

    3) Find the distance of the point (3,5) from the line 2x+3y=14 measured parallel to a line having slope 1/2.[Book Answer:5^(1/2)](Solved)

    4) Find the distance of the point (2,5) from the line 3x+y+4=0 measured parallel to a line having slope 3/4.[Book Answer: 5 units](Solved)

    That's all. Thanks
    Last edited by shubhadeep; December 30th 2005 at 09:25 AM.
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  2. #2
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    Quote Originally Posted by shubhadeep
    Hi I have problems in some questions.
    A) My answers to these questions do not match the book answers. It can be a misprint though:-

    1) The three vertices of a paralleologram are (3,4), (3,8) and (9,8). Find the fourth vertex.[Book Answer9,4)]
    The book is right. Get some squared paper and draw the given points in;
    you will see that the answer is obvious (the lines through them meet at a
    right angle, so your parallelogram is a rectangle).

    RonL
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  3. #3
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    thanks for the second question. i got it. since the area is in the form of a modulus, therefore there should have been two values for the area. i had put only 5 and not -5. that was my mistake.
    still trying the first one.
    Last edited by shubhadeep; December 30th 2005 at 01:35 AM.
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  4. #4
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    Correction for Q2

    Quote Originally Posted by shubhadeep

    2) The area of a triangle is 5. Two of its vertices are (2,1) and (3,-2). The third vertex lies on y=x+3. Find the third vertex.[Book Answer: (7/2,13/2) and (-3/2,3/2). How am I supposed to get the second answer]
    The second answer is the point on the given line on the other side of the
    line through (2,1) and (3,-2) from the first solution which gives the required
    area. (Do a sketch and you will see that these are both solutions).

    RonL
    Last edited by CaptainBlack; December 30th 2005 at 03:41 AM. Reason: Original was in error
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  5. #5
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    Quote Originally Posted by CaptainBlack
    The bisector of angle A in triangle ABC is the line throught
    point A, and the mid-point of segment BC.
    But how do u know that the angle bisectorof A bisects BC.
    I still tried it. The equation I'm getting is x-2y+2=0.
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  6. #6
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    Quote Originally Posted by shubhadeep
    But how do u know that the angle bisectorof A bisects BC.
    I still tried it. The equation I'm getting is x-2y+2=0.
    You are right usually it doesn't

    You need to find the point D on segment AB (produced if need be) a distance
    AC from A, then use the mid point of segment CD not the midpoint of BC -
    but this looks too fussy there must be a simpler way

    Sorry about that - I appear not to be on form today.

    RonL
    Last edited by CaptainBlack; December 30th 2005 at 03:42 AM.
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  7. #7
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    Quote Originally Posted by shubhadeep
    3) Find the distance of the point (3,5) from the line 2x+3y=14 measured parallel to a line having slope 1/2.[Book Answer:5^(1/2)]
    Find the equation of the line through (3,5) of slope 1/2.

    Then find the point P at which this line meets 2x+3y=14.

    Finally find the distance between (3,5) and P.

    RonL
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  8. #8
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    Quote Originally Posted by CaptainBlack
    You need to find the point D on segment AB (produced if need be) a distanceAC from A, then use the mid point of segment CD not the midpoint of BC -but this looks too fussy there must be a simpler way.
    Yea, there has to be a simpler way. I did try it by the method you say earlier but I got stuck. Will try again.

    Sorry about that - I appear not to be on form today.
    No Problem. Happens to many.
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  9. #9
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    Quote Originally Posted by shubhadeep
    2) Find the equation of the bisector of angle A of the triangle whose vertices are A(4,3), B(0,0) and C(2,3).[Book Answer: x-3y+5=0]
    Here we go again. Find the point D on AB the same distance as C
    is from A. As B is (0,0) this is fairly simple to do, |BA|=5 so simple
    proportion will give D is the point \frac{3}{5}(4,3)=(2.4,1.8).

    Then the mid point of CD is (2.2,2.4).

    So the bisector is the line through (4,3) and (2.2,2.4).

    So not as fussy as I thought.

    RonL
    Attached Thumbnails Attached Thumbnails Some problems-bisector.jpg  
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  10. #10
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    Yeah Thanks! I never thought of that.
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  11. #11
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    As B is (0,0) this is fairly simple to do, |BA|=5 so simple proportion will give D is the point \frac{3}{5}(4,3)=(2.4,1.8).

    I am not familiar with using proportions in co-ordinates. Can you explain it please. And how would I have done that if the point B was not on the origin.
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  12. #12
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    Quote Originally Posted by shubhadeep
    I am not familiar with using proportions in co-ordinates. Can you explain it please. And how would I have done that if the point B was not on the origin.
    BA has length 5 (since BA(4,0) is a 3,4,5 right triangle).
    So we want a point on D on BA such that |DA|=2, and |BD|=3. So if D is
    (d_1,d_2) then triangle BD(d_1,0) is similar to BA(4,0), with hypotenuses
    in the ratio of 3:5. So the heights are in the same ratio, so d_2=(3/5)3.
    The same applies to the horizontal legs so d_1=(3/3)4.

    RonL
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  13. #13
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    Quote Originally Posted by shubhadeep

    3) The point A divides the join of P(-5,1) and Q(3,5) in the ratio k:1. Find the values of k for which the area of triangle ABC where B is (1,5) and C(7,-2) is equal to 2 units.[Book Answer: k=7 or 31/9]
    I hope you will find this of help in solving this problem:

    As A lies on the line on which P and Q it can be written:

    A=(1-\lambda).P+\lambda.Q

    for some \lambda \epsilon \mathbb{R} (here we are treating A, P and Q as the 2-vectors of their coordinates).

    The distance of A from P is:

    D_{AP}=|\lambda|.|P-Q|,

    and the distance of A from Q is:

    D_{AQ}=|1-\lambda|.|P-Q|.

    Now the condition that the point A divides the join of P(-5,1)
    and Q(3,5) in the ratio k:1 is equivalent to:

    \frac{D_{AP}}{D_{AQ}}=k

    so:

    \frac{|\lambda |}{|1-\lambda|}=k.

    Which should be sufficient information to allow you to find A in terms of k
    and the coordinates of P and Q.

    I presume that you know the formula for the area of a triangle in terms of the
    coordinates of its vertices, or failing that Heron's formula for the area of a
    triangle, and so should be able to complete this problem from here.

    RonL
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