×

# 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.

3 years, 5 months ago

Sort by:

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. · 3 years, 5 months ago

I will corroborate your solution there. · 3 years, 5 months ago

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. · 3 years, 5 months ago

there's a easy method using determinant criteria, for 3 points prove that the determinant of area equal to 0 · 3 years, 5 months ago

This method is far more convenient in exams, since it would take much less time. Determinant method is always applicable to show collinearity. · 3 years, 5 months ago

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'. · 3 years, 5 months ago

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 $$x$$. · 3 years, 5 months ago

at x=-1........-x/2 and cos(x) are same.............so Domain,I guess, is R-{-1}...... · 3 years, 5 months ago

Comment deleted May 21, 2013

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. · 3 years, 5 months ago

then I don't think it can be onto. · 3 years, 5 months ago

$$cos 1$$ is 0.540, not exactly 0.5. · 3 years, 5 months ago

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?? · 3 years, 5 months ago

the co-domain is real numbers? · 3 years, 5 months ago

hey,you gave isi bmath exam,right...? · 3 years, 5 months ago