question_answer

If the sum of the distances of a point from two perpendicular lines in a plane is 1, then its locus is [IIT 1992, Karnataka CET 1999; DCE 2000,01]

A)            Square                                       

B)            Circle

C)            Straight line                              

D)            Two intersecting lines

Correct Answer: A

Solution :

           Required locus of the point \[(x,y)\] is the curve\[|x|+|y|=1\]. If the point lies in the first quadrant, then \[x>0,y>0\] and so \[|x|+|y|=1\Rightarrow x+y=1\], which is straight line AB. If the point \[(x,\,y)\]lies in second quadrant then \[x<0\], \[y>0\] and so \[|x|+|y|=1\] Þ \[-x+y=1\] Similarly for third and fourth quadrant, the equations are \[-x-y=1\]and \[x-y=1\].                    Hence the required locus is the curve consisting of the sides of the square ABCD.