question_answer

In the rectangular co-ordinate system above, if the equation of \[{{l}_{1}}\] is y = x and parallel to \[{{l}_{2}}\], the shortest distance between \[{{l}_{1}}\] and \[{{l}_{2}}\]is

A)  \[\sqrt{2}\]                                       

B)  1

C)  \[\frac{\sqrt{2}}{2}\]                      

D)         \[\frac{1}{2}\]

Correct Answer: C

Solution :

 Here, equation of line \[{{l}_{2}}\] is,  \[y=x\]. \[\therefore \]  Slope of \[{{l}_{1}}=\frac{x}{y}=1.\]i.e. \[{{l}_{1}}\] makes \[{{45}^{o}}\] angle with both x-axis and y- axis. Now, we are asked to find the shortest distance between \[{{l}_{1}}\] and \[{{l}_{2}}\] which will be the perpendicular distance between \[{{l}_{1}}\] and \[{{l}_{2}}\] . So, drop a perpendicular LM as shown below:                 Now, in \[\Delta \,LMO,\,\angle M={{90}^{o}}\] and        \[MOL={{45}^{o}}\] \[\therefore \]  \[\angle OLM={{180}^{o}}-(\angle M+\angle MOL)\]                                 \[={{180}^{o}}-({{90}^{o}}+{{45}^{o}})={{45}^{o}}\] So, \[\Delta \,LMO\]is isosceles right triangle. Now, in \[\Delta \,LMO,\] hypotenuse, \[LO=1\]. \[\therefore \]                  \[LM=\frac{1}{\sqrt{2}}\] (Length of hypotenuse)                 \[=\frac{1}{\sqrt{2}}(1)=\frac{1}{\sqrt{2}}\times \frac{\sqrt{2}}{\sqrt{2}}=\frac{\sqrt{2}}{2}\] So, the shortest distance between \[{{l}_{1}}\] and \[{{l}_{2}}\]is \[\frac{\sqrt{2}}{2}\].