2010
Sat
6
Mar
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M Is For Miracle

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Posted at 8:45 pm by Miss Loi in A-Maths Questions

Continuing from last week’s episode

Night has fallen when the Maths Tutor reached her car.

Just as she was about to start the engine and drive away, the street lamps began to flicker wildly.

A great sense of foreboding filled the air, as a strange vision suddenly flashed before her eyes …

In an imaginary grid, a man walks along a straight street defined by the equation: mx + (m − 1)y = 10

At the same time, a black car zooms furiously along a straight road defined by the equation: (m − 2)x + 3my = 20

The car will hit the man if their paths meet …

A scream snapped her out of her trance before her vision was complete, and she blinked her eyes in time to see a frightened woman pushed a man away under the glare of the bright street lamps.

The man (who happens to be our young henchman if you’ve been following the previous episodes) then turned around and walked away with his head bowed, looking dejected.

In a shocking realization of her déjà vu experience, she recognized the man as the walking man in her vision just moments ago.

And in that very moment, the silence of the night was broken by the unmistakable roar of a speeding car’s engine, while the Maths Tutor could only cover her eyes and pray for a miracle …

find the ‘miracle’ value of m where the paths of the car and the man would never meet.
頑張って!!!

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11 Comments

  1. e.lee's Avatar
    e.lee says


    2010
    Mar
    8
    Mon
    4:13am
     
    1

    do students still have problems with English and General Paper? gawwd its been sooooo loonng since I've done my exams.

  2. clarion-x's Avatar
    clarion-x says


    2010
    Mar
    9
    Tue
    8:58am
     
    2

    From eqn 1, mx + (m -1)y = 10
    mx+my-y=10
    10-mx=my-y
    y=(10-mx)/(m-1)

    From eqn 2, (m -2)x + 3my = 20
    mx-2x+3my=20
    y=(20+2x-mx)/(3m)
    hence, (20+2x-mx)/(3m)=(10-mx)/(m-1)
    cross multiplying,
    30m-3m^2x=20m+2mx-m^2x-20-2x+mx
    30m-3m^2x-20m-2mx-m^2x+20+2x-mx=0
    10m-4m^2x-3mx+20+2x=0
    -4m^2x-(3m-2)x+(10m+20)=0
    4m^2x+(3m-2)x-(10m+20)=0

    If both paths don't meet, b^2-4ac<0
    b^2-4ac=(3m-2)^2-4(4m^2)(-10m-20)
    simplifying,... (lazy to type out here)
    m less than -1.99? O: i think its wrong leh...

  3. mathslover's Avatar
    mathslover says


    2010
    Mar
    9
    Tue
    11:51am
     
    3

    @clarion-x,
    The 2nd step after cross multiplying, when you moved m^2x over you forgot to change sign. Also, when you use b^2-4ac your equation must be quadratic. Perhaps you have overlooked the fact there is no x^2 term?
    _________________________________________
    Equation 1:
    mx + (m-1)y = 10
    {10-mx}/{m-1} = y
    y=(m/{1-m})x+10/{m-1}

    Equation 2:
    (m-2)x+3my=20
    3my=20-(m-2)x
    y=20/{3m} - {(m-2)x}/{3m}

    Both equations are linear. For two linear equations to not cross paths, they must be parallel, and have different y-intercepts.

    gradient of equation 1 = gradient of equation 2
    m/{1-m} = {2-m}/{3m}
    3m^2=(1-m)(2-m)
    2m^2+3m-2=0
    (m+2)(2m-1)=0
    m=-2 or m=1/2

    When m=-2,
    y intercept of equation 1 = 10/{m-1} = 10/-3
    y intercept of equation 2 = 20/3m = 20/-6 = 10/-3
    Hence m cannot be -2.

    When m={1/2}
    y intercept of equation 1 = {10/{m-1}} = {10/{-1/2}} = -20
    y intercept of equation 2 = 20/3m = {20/{3/2}} = 40/3
    Hence m={1/2}.

    The speeding roar of the car shocked the henchman to a standstill, he stared for 0.5 seconds at the oncoming car, and did the unthinkable...

  4. clarion-x's Avatar
    clarion-x says


    2010
    Mar
    9
    Tue
    8:50pm
     
    4

    Argh <.<

    ...He leapt up gently onto the branch of a tall Angsana tree by the road with a level of 轻功 that far surpassed any of the characters in the TVB 古装剧 that Miss Loi faithfully followed.
    In a flash the henchman had disappeared, leaving Miss Loi's mouth agape in the shape of x^2+y^2=1. What was that? She thought. It was no where near the Seventh Month...

    P.S: I realise I've forgotten a lot of my O level Maths, but I haven't forgot my story-spinning skills! :D

  5. mathslover's Avatar
    mathslover says


    2010
    Mar
    9
    Tue
    11:28pm
     
    5

    ... with trembling fingers, Miss Loi quickly started her car, speeding down the road defined by the equation -5x - y = 40. The henchman, watching with eagle eyes from the angsana tree beside the road, noticed that this was the car his boss mentioned.

    Glancing down, he quickly pictured the road to be positioned along -5x - y = 40, on the scale of 1 unit = 1 metre.

    Miss Loi's car (when she stared with her mouth open in the shape of x^2+y^2=1) was situated on (-10,10) and the Angsana tree was situated at the origin. The Angsana tree was 2m tall.

    If Miss Loi accelerated at a constant rate of 2m/s²,find the time t after Miss Loi started her car, that the henchman has to leap off the angsana tree, so that he will land on her car.

    Let the acceleration of free fall be 10m/s²...

    And then Miss Loi's eyebrows started moving into the shape of y = |x| as she returned to discover two overaged students hijack her question until 体无完肤...



  6. 2010
    Mar
    10
    Wed
    12:47am
     
    6

    Welcome to Jφss Sticks e.lee! While Miss Loi may not be that much into the English/GP scene, she's sure that they remain the bane of a fair amount of students.

    In fact this may well be the Achilles' Heel of the formidable foreign cyborgs studying amongst us!

    clarion + mathslover,

    This is taking 無厘頭 to the extreme!!!

    ... on the scale of 1 unit = 1 metre ... she stared with her mouth open in the shape of x² + y² = 1

    So the radius of Miss Loi's wide open mouth is 1 m???

    *FACEPALM*

  7. clarion-x's Avatar
    clarion-x says


    2010
    Mar
    10
    Wed
    7:07pm
     
    7

    *Looks at mathlover* No lah Miss Loi's mouth 10cm in radius niah! xD



  8. 2010
    Mar
    10
    Wed
    7:40pm
     
    8

    *pulls everyone back to the pre-Angsana tree time frame as we have yet to finish our discussion on the original question!*

    As the roar of the car drew nearer, the Maths Tutor closed her eyes in fear, and another vision appeared to her ... a vision of a series of similar questions that has been appearing in some of her students' tests and exams, asking them to determine if there's an intersection, a solution, common solution, unique solution etc. to a pair of linear equations/lines.

    Regardless of the phrase used, the scenarios of all kinds of 'solutions' between a pair of simultaneous equations l1 and l2 can be summarised in three cases as follows:

    Solutions of Linear Equations

    In our question, we are looking for the value of m such that the two non-overlapping lines don't meet i.e. in mathslover's words

    they must be parallel, and have different y-intercepts

    which is the 'No Solutions' (Case 2(a)) described in the diagram above.

    To obtain this value of m, one may either begin with equating the gradients of both lines as in mathlover's working or via the determinant of the coefficient matrix in an associated matrix equation (see below).

    No matter which method is used, you'll almost certainly end up with a quadratic equation with two values of m (one for each scenario (a) & (b) under Case 2) since being parallel only means that there's no unique solution (or intersection point) but both equations can still belong to the same line with ∞ solutions (Case 2(b)).

    ∴ you'll need to substitute the values of m obtained into both equations (as shown in the second part of mathlover's working in order to ascertain the final value of m = 1/2 that yields two parallel lines with no common solution as required by the question.

    Note that clarion's substitution approach won't work because we're finding m (instead of x and y) and this will lead to a single equation with two unknowns (m and x).


    As an added note, this kind of question usually forms part of a problem that explicitly requires one to solve a pair of simultaneous linear equations using the inverse matrix method.

    With this approach, we can convert our two equations:

    mx + (m − 1)y = 10
    (m − 2)x + 3my = 20

    to matrix form:

    delim{[}{matrix{2}{1}{{mx+(m-1)y} {(m-2)x+3my}}}{]} = delim{[}{matrix{2}{1}{10 20}}{]}
    doubleright delim{[}{matrix{2}{2}{m m-1 m-2 {3m}}}{]} delim{[}{matrix{2}{1}{x y}}{]} = delim{[}{matrix{2}{1}{10 20}}{]}

    In order for the above to have no unique solution, the coefficient matrix delim{[}{matrix{2}{2}{m m-1 m-2 {3m}}}{]} must be singular i.e. its determinant = 0.

    So determinant of delim{[}{matrix{2}{2}{m m-1 m-2 {3m}}}{]} = delim{|}{matrix{2}{2}{m m-1 m-2 {3m}}}{|} = m(3m)-(m-1)(m-2) = 0

    Expanding the above we have our quadratic equation for m:
    2m² + 3m − 2 = 0
    (m − 1/2)(m + 2) = 0
    m = 1/2 or −2

    Sub m=1/2 into the equations:
    Eqn 1: (1/2)x − 1/2y = 10 ⇒ y = x − 20
    Eqn 2: (−3/2)x + 3/2y = 20 ⇒ y = x + 40/3
    ⇒ two parallel lines with different y-intercepts

    Sub m=−2 into the equations:
    Eqn 1: −2x − 3y = 10
    Eqn 2: −4x − 6y = 20 ⇒ −2x − 3y = 10
    ⇒ Eqn 2 is the same as Eqn 1!

    m = 1/2 if the two lines don't meet (no common solution).


    Now to decide what to do with the ridiculous 轻功 and angsana tree that has suddenly sprung up ...

  9. mathslover's Avatar
    mathslover says


    2010
    Mar
    11
    Thu
    12:05am
     
    9

    *Looks at clarion*: Miss Loi's mouth is apparently 14.1m away from her body...

    shape of x^2+y^2=1 != location of x^2+y^2=1 != size of x^2+y^2=1 !=...

    But if so desired, Miss Loi's mouth will be open in the shape of (x+10)^2+(y-10)^2=1, at the same scale. :D

    And to be honest I've always thought of matrices as a method used to solve large numbers of simultaneous linear equations, via RREF. For 2 equations isn't this a bit 杀鸡用牛刀?



  10. 2010
    Mar
    11
    Thu
    12:47am
     
    10

    A monster with a 1 m radius mouth and fierce v-shaped y = |x| eyebrows appears suddenly and speaks to mathslover:

    With the migration of the majority of the Matrices topic from the old A-Maths 4018 Syllabus to the current E-Maths 4016 Syllabus, the only trace of matrices that's left in the current 4038 AMaths syllabus is the part that deals with expressing a pair of linear equations in matrix form and solving the equations by inverse matrix method, which basically covers concepts like inverse matrices, determinants, singular/non-singular matrices etc. - stuff that have been left out of EMaths.

    As such, it is likely that one will be told and compelled to 杀 two simultaneous equations 用 inverse matrix method within the question, in order for the examiner to cover this portion of the syllabus in your AMaths paper.

    Anyway it's a pretty fast, one-step process to form the quadratic equation for m using the determinant right?

  11. mathslover's Avatar
    mathslover says


    2010
    Mar
    11
    Thu
    11:32pm
     
    11

    Staring into the 1 m radius mouth, she answered:

    So it's one of those 'hence' questions that we cannot use the 'otherwise' method to solve?

    In most 'hence or otherwise' questions, the 'hence' method is usually the obvious and straightforward one. It is easier to use the determinant to solve in this case, but not so easy to grasp the concept of why or how it works.

    *wonders aloud how matrices are actually taught in secondary schools... students nowadays are expected to know so much more, sigh*

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    Miss Loi tweeted


    2010
    Mar
    6
    Sat
    2:33pm
     
    12

    M is for Miracle: http://bit.ly/bDN6tS

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