question_answer

Using section formula, check that the points A (- 3, -1), B (1, 3) and C (-1, 1) are collinear.

A) Yes

B) No

C) Can't say

D) None of these

Correct Answer: A

Solution :

Let \[C\left( -1,\,\,1\, \right)\]divides AB in the ratio k : 1.

Then, by using section formula, we get

Coordinates of C are \[\left( \frac{k-3}{k+1},\,\frac{3k-1}{k+1} \right)\].

Thus, \[C\left( -1,\,1 \right)=C\left( \frac{k-3}{k+1},\,\frac{3k-1}{k+1} \right)\]

On equating  x-coordinate from both sides, we get

\[-1=\frac{k-3}{k+1}\Rightarrow \,\,\,-k-1=k-3\]

\[\Rightarrow \,\,\,-2k=-3+1\Rightarrow \,\,-2k=-2\Rightarrow \,k=1\]

On equating y-coordinate from both sides, we get

\[1=\frac{3k-1}{k+1}\]

\[\Rightarrow \,\,\,k+1=3k-1\]

\[\Rightarrow \,\,\,2k=2\,\,\Rightarrow \,\,\,k=1\]

Since, in both cases value of k is same.

So, C divides AB in the ratio 1:1, i.e. C is the mid-point of AB.

Hence, A, B and C are collinear.