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.You need to login to perform this action.
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