Collinearity

Let a1,a2,a3,....ana_{1}, a_{2}, a_{3},....a_{n} and b1,b2,b3,...bnb_{1} , b_{2}, b_{3},...b_{n} be two arithmetic progressions.

Then prove that the points : (a1,b1),(a2,b2),.......,(an,bn) (a_{1},b_{1} ),(a_{2},b_{2}),.......,(a_{n},b_{n}) are collinear.

Note by Aditya Parson
6 years, 5 months ago

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We will now show that any 3 points are collinear.

Thus, we have to show (ak,bk)(a_k, b_k), (am,bm)(a_m, b_m) and (an,bn)(a_n, b_n) are collinear.

    \iff Line formed by (ak,bk)(a_k, b_k) and (am,bm)(a_m, b_m) is parallel to line formed by (am,bm)(a_m, b_m) and (an,bn)(a_n, b_n)

Using the fact of arithmetic progressions, we can let ak=a0+k×daa_k=a_0+k \times d_a and bk=b0+k×dbb_k=b_0+k \times d_b, for some constants a0,da,b0,dba_0, d_a, b_0, d_b

Therefore we can show that the gradients of the 2 lines are equal.

Gradient of line formed by (ak,bk)(a_k, b_k) and (am,bm)=bmbkamak=dbda(a_m, b_m)=\frac{b_m-b_k}{a_m-a_k}=\frac{d_b}{d_a}

Gradient of line formed by (am,bm)(a_m, b_m) and (an,bn)=bmbnaman=dbda(a_n, b_n)=\frac{b_m-b_n}{a_m-a_n}=\frac{d_b}{d_a}

Since gradients are equal, it follows that the lines are parallel and thus (ak,bk)(a_k, b_k), (am,bm)(a_m, b_m) and (an,bn)(a_n, b_n) are collinear.

Now we can start with base case n=3 and induct by taking points (a1,b1),(a2,b2)(a_1, b_1), (a_2, b_2) and (an+1,bn+1)(a_{n+1}, b_{n+1}). Hence proven.

Clarence Chew - 6 years, 5 months ago

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I will corroborate your solution there.

Aditya Parson - 6 years, 5 months ago

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Let ai=λai+μaa_i = \lambda_a i+\mu_a, bi=λbi+μbb_i = \lambda_b i+\mu_b for some real numbers λa,λb,μa,μb\lambda_a,\lambda_b,\mu_a,\mu_b.

Then i=aiμaλai=\frac{a_i-\mu_a}{\lambda_a}, and thus bi=λbaiμaλa+μb=λbλaai+(μbμaλbλa)b_i = \lambda_b\frac{a_i-\mu_a}{\lambda_a}+\mu_b = \frac{\lambda_b}{\lambda_a}a_i+(\mu_b-\frac{\mu_a\lambda_b}{\lambda_a}).

Hence all the points (ai,bi)(a_i,b_i) lie on the same line y=λbλax+(μbμaλbλa)y=\frac{\lambda_b}{\lambda_a}x+(\mu_b-\frac{\mu_a\lambda_b}{\lambda_a}), as desired.

Ang Yan Sheng - 6 years, 5 months ago

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This prob came in C.M.I....................... Do you know how to prove that f(x)=1/(x+2cos(x)) is onto rigorously...I mean intuitively it can be seen, but is there any othere way to do it.............I mean using 'hardcore math'.

Piyal De - 6 years, 5 months ago

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There exists f(x)f(x) whenever x+2cosx=0x+2cosx \neq=0. But I don't know how can we conclude that xx will never be =2cosx=-2cosx for all xx.

Aditya Parson - 6 years, 5 months ago

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at x=-1........-x/2 and cos(x) are same.............so Domain,I guess, is R-{-1}......

Piyal De - 6 years, 5 months ago

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hey,you gave isi bmath exam,right...?

Sayan Chaudhuri - 6 years, 5 months ago

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Nope friend, B.Stat............what about you?

Piyal De - 6 years, 5 months ago

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there's a easy method using determinant criteria, for 3 points prove that the determinant of area equal to 0

Pratik Singhal - 6 years, 5 months ago

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This method is far more convenient in exams, since it would take much less time. Determinant method is always applicable to show collinearity.

Aditya Parson - 6 years, 5 months ago

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