question_answer

The intercepts on the straight line \[y=mx\] by the lines y = 2 and y = 6 is less than 5, then m belongs to

A) \[\left] -\frac{4}{3},\frac{4}{3} \right[\]                 

B) \[\left] \frac{4}{3},\frac{3}{8} \right[\]

C)  \[\left. \left] -\infty ,\frac{-4}{3}, \right[\cup  \right]\frac{4}{3},\infty \left[ \begin{align}   &  \\  &  \\ \end{align} \right.\]

D) \[\left] \frac{4}{3},\infty  \right[\]

Correct Answer: C

Solution :

Give lines are \[y=mx,y=2\]and \[y=6\] \[\therefore \]Coordinates of points A and B are \[\left( \frac{2}{m},2 \right)\]and \[\left( \frac{6}{m},6 \right)\], respectively. \[\therefore \]\[AB=\sqrt{{{\left( \frac{2}{m}-\frac{6}{m} \right)}^{2}}+{{(2-6)}^{2}}<5}\] \[\Rightarrow \]               \[{{\left( \frac{2}{m}-\frac{6}{m} \right)}^{2}}+{{4}^{2}}<25\] \[\Rightarrow \]               \[{{\left( \frac{2}{m}-\frac{6}{m} \right)}^{2}}<9\] \[\Rightarrow \]               \[-3<\frac{2}{m}-\frac{6}{m}<3\] \[\Rightarrow \]               \[-3<-\frac{4}{m}<3\Rightarrow -\frac{4}{3}>m\frac{4}{3}\] \[\therefore \]  \[m\in \left. \left] -\infty ,\frac{-4}{3} \right[\cup  \right]\frac{4}{3},\infty \left[ _{_{_{{}}^{{}}}^{{}}}^{{}} \right.\]