A) \[1,2\]
B) \[2,\frac{1}{2}\]
C) \[-\frac{1}{2},2\]
D) \[-\,\,2,\frac{1}{2}\]
Correct Answer: B
Solution :
| Since, the points \[A\,\,(k+1,2k),\]\[B\,\,(3k,2k+3)\]and |
| \[C\,\,(5k-1,5k)\] are collinear. |
| Area of \[\Delta ABC=0\] |
| \[\frac{1}{2}[{{x}_{1}}({{y}_{2}}-{{y}_{3}})+{{x}_{2}}({{y}_{3}}-{{y}_{1}})+{{x}_{3}}({{y}_{1}}-{{y}_{2}})]=0\] |
| \[\Rightarrow \] \[\frac{1}{2}[(k+1)(2k+3-5k)+3k(5k-2k)\]\[+\,\,(5k-1)(2k-(2k+3)]=0\] |
| \[\Rightarrow \] \[\frac{1}{2}[(k+1)(-3k+3)+3k(3k)\] |
| \[+(5k-1)(2k-2k-3)]=\] |
| \[\Rightarrow \] \[\frac{1}{2}[-3{{k}^{2}}+3k-3k+3+9{{k}^{2}}-15k+3]=0\] |
| \[\Rightarrow \] \[\frac{1}{2}[6{{k}^{2}}-15k+6]=0\] (Multiply by 2) |
| \[\Rightarrow \] \[6{{k}^{2}}-15k+6=0\] |
| \[\Rightarrow \] \[2{{k}^{2}}-5k+2=0\] (Divide by 3) |
| \[\Rightarrow \] \[2{{k}^{2}}-4k-k+2=0\] |
| \[\Rightarrow \]\[2k\,\,(k-2)-1\,\,(k-2)=0\] |
| \[\Rightarrow \] \[(k-2)(2k-1)=0\] |
| If \[k-2=0\]\[\Rightarrow \]\[k=2\] |
| If \[2k-1=0\]\[\Rightarrow \]\[k=\frac{1}{2}\]\[\therefore \]\[k=2,\frac{1}{2}\] |
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