A) \[\frac{1}{h}+\frac{1}{k}=\frac{1}{3}\]
B) \[\frac{1}{k}-\frac{1}{h}=\frac{1}{3}\]
C) \[h+k=3\]
D) \[\frac{1}{h}-\frac{1}{k}=\frac{1}{3}\]
Correct Answer: A
Solution :
[a] \[\therefore (3,3),(h,0)\And (0,k)\]be collinear. \[\because {{R}_{2}}\to {{R}_{2}}-{{R}_{2}}\] \[{{R}_{3}}\to {{R}_{3}}-{{R}_{1}}\] \[h(k-3)-3k=0\] \[hk-3h-3k=0\] \[hk=3k+3h\] \[1=\frac{3}{h}+\frac{3}{k}\Rightarrow \frac{1}{3}=\frac{1}{h}+\frac{1}{h}\]You need to login to perform this action.
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