question_answer

The locus of a point so that sum of its distance from two given perpendicular lines is equal to 2 unit in first quadrant, is  [Bihar CEE 1994]

A)            \[x+y+2=0\]                               

B)            \[x+y=2\]

C)            \[x-y=2\]                                   

D)            None of these

Correct Answer: B

Solution :

           We take the coordinate axes as two perpendicular lines. Let \[P\,({{x}_{1}},{{y}_{1}})\] be the required point.                    From \[P\,({{x}_{1}},{{y}_{1}})\], we draw PM and PN perpendicular to OX and OY respectively.            Given, \[PM+PN=2\]                ......(i)                    But,     \[PM={{y}_{1}},PN={{x}_{1}}\]                    Hence from (i), \[{{y}_{1}}+{{x}_{1}}=2\]                    Thus locus of \[({{x}_{1}},{{y}_{1}})\]is \[x+y=2\]                    which is a straight line.