| If \[P\,\,(a,0),\]\[Q\,\,(0,b)\] and \[R\,\,(1,1)\] are collinear, then find the value of \[\frac{1}{a}+\frac{1}{b}\] |
A) \[2\]
B) \[1\]
C) \[-\,\,1\]
D) \[0\]
Correct Answer: B
Solution :
| Since, P, Q and R are collinear |
| \[\therefore \]Area of \[\Delta PQR=0,\] |
| then \[\frac{1}{2}[{{x}_{1}}({{y}_{2}}-{{y}_{3}})+{{x}_{2}}({{y}_{3}}-{{y}_{1}})+{{x}_{3}}({{y}_{1}}-{{y}_{2}})]=0\] |
| \[\therefore \]\[{{x}_{1}}=a,\]\[{{x}_{2}}=0,\]\[{{x}_{3}}=1,\]\[{{y}_{1}}=0,\]\[{{y}_{2}}=b\]and \[{{y}_{3}}=1\] |
| \[\Rightarrow \]\[\frac{1}{2}[a\times (b-1)+0\times (1-0)+1\times (0-b)]=0\] |
| \[\Rightarrow \]\[\frac{1}{2}[ab-b-a]=0\]\[\Rightarrow \]\[ab=a+b\] |
| \[\Rightarrow \]\[1=\frac{a+b}{ab}\]\[\Rightarrow \]\[\frac{1}{a}+\frac{1}{b}=1\] |
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