question_answer

The points \[\left( \mathbf{3},\mathbf{3} \right),\left( \mathbf{h},\mathbf{0} \right)\]and \[\left( \mathbf{0},\mathbf{k} \right)\]are collinear if

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. \[\because {{R}_{2}}\to {{R}_{2}}-{{R}_{2}}\] \[{{R}_{3}}\to {{R}_{3}}-{{R}_{1}}\]   \[h(k-3)-3k=0\] \[hk-3h-3k=0\] \[hk=3k+3h\] \[1=\frac{3}{h}+\frac{3}{k}\Rightarrow \frac{1}{3}=\frac{1}{h}+\frac{1}{h}\]