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. |
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