Question

If the sum of the distances of a moving point from two perpendicular lines in a plane is always 1 then its locus is

Answer 1

Hey.

Here is the answer.

Let the perpendiculars lines be the x – axis and y – axis.  The sum of the distances from the the point P (x, y) is 1. ie.

|x| + |y| = 1  – this is the locus of the point P which would be the rhombus whose sides are 

x + y = 1; –x + y = 1; x – y = 1; –x – y = 1

If the perpendicular lines are other than co-ordinate axes, then we can get the same solution by using the transformation of the axes concept.

So , rhombus or can say square is the answer.

Thanks.

Answer 2

Consider that the perpendicular lines are chosen along the axes and if (x,y) is the point in the first quadrant, then the equation will be,

x+y=1

In the same way, if the point (x,y) lies in second, third and fourth quadrant, then the equations will be,

x−y=1,−x+y=1,−x−y=1

This shows that,

x+y=±1 and x−y=±1.

It can be clearly observed that the above four lines encloses a square.