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 \[\frac{\sqrt{2}}{2}\].You need to login to perform this action.
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