Waste less time on Facebook — follow Brilliant.
×

Collinearity

Let \(a_{1}, a_{2}, a_{3},....a_{n}\) and \(b_{1} , b_{2}, b_{3},...b_{n}\) be two arithmetic progressions.

Then prove that the points : \( (a_{1},b_{1} ),(a_{2},b_{2}),.......,(a_{n},b_{n}) \) are collinear.

Note by Aditya Parson
4 years, 5 months ago

No vote yet
6 votes

Comments

Sort by:

Top Newest

We will now show that any 3 points are collinear.

Thus, we have to show \((a_k, b_k)\), \((a_m, b_m)\) and \((a_n, b_n)\) are collinear.

\(\iff\) Line formed by \((a_k, b_k)\) and \((a_m, b_m)\) is parallel to line formed by \((a_m, b_m)\) and \((a_n, b_n)\)

Using the fact of arithmetic progressions, we can let \(a_k=a_0+k \times d_a\) and \(b_k=b_0+k \times d_b\), for some constants \(a_0, d_a, b_0, d_b\)

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

Gradient of line formed by \((a_k, b_k)\) and \((a_m, b_m)=\frac{b_m-b_k}{a_m-a_k}=\frac{d_b}{d_a}\)

Gradient of line formed by \((a_m, b_m)\) and \((a_n, b_n)=\frac{b_m-b_n}{a_m-a_n}=\frac{d_b}{d_a}\)

Since gradients are equal, it follows that the lines are parallel and thus \((a_k, b_k)\), \((a_m, b_m)\) and \((a_n, b_n)\) are collinear.

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

Clarence Chew - 4 years, 5 months ago

Log in to reply

I will corroborate your solution there.

Aditya Parson - 4 years, 5 months ago

Log in to reply

Let \(a_i = \lambda_a i+\mu_a\), \(b_i = \lambda_b i+\mu_b\) for some real numbers \(\lambda_a,\lambda_b,\mu_a,\mu_b\).

Then \(i=\frac{a_i-\mu_a}{\lambda_a}\), and thus \(b_i = \lambda_b\frac{a_i-\mu_a}{\lambda_a}+\mu_b = \frac{\lambda_b}{\lambda_a}a_i+(\mu_b-\frac{\mu_a\lambda_b}{\lambda_a})\).

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

Ang Yan Sheng - 4 years, 5 months ago

Log in to reply

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

Pratik Singhal - 4 years, 5 months ago

Log in to reply

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 - 4 years, 5 months ago

Log in to reply

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 - 4 years, 5 months ago

Log in to reply

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

Aditya Parson - 4 years, 5 months ago

Log in to reply

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

Piyal De - 4 years, 5 months ago

Log in to reply

Comment deleted May 21, 2013

Log in to reply

@Aditya Parson Buddy, onto means that f(x) will take all values in its range for x>=0, right? You showed that deno can never be = to zero for x>2/x<-2, so how can we say that it's onto from that..........as for equality, check the graph at x=-1, they're =. Actually the original prob was to find the range of f(x)=1/(x+2cos(x)), but I was sort of trying to solve it in another way so i asked the question.

Piyal De - 4 years, 5 months ago

Log in to reply

@Piyal De then I don't think it can be onto.

Aditya Parson - 4 years, 5 months ago

Log in to reply

@Piyal De \( cos 1\) is 0.540, not exactly 0.5.

Aditya Parson - 4 years, 5 months ago

Log in to reply

@Aditya Parson Well actually I didn't do it..........but check the graph on a graphing cal.you'll see it as clear as crystal. Wat abt onto??

Piyal De - 4 years, 5 months ago

Log in to reply

@Piyal De the co-domain is real numbers?

Aditya Parson - 4 years, 5 months ago

Log in to reply

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

Raja Metronetizen - 4 years, 5 months ago

Log in to reply

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

Piyal De - 4 years, 5 months ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...