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Question:
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If the sum of the distances of a moving point in a plane from the axes is 1 , then find the locus of the point.
Solution:
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Verified Answer
Let the coordinates of moving point be $(\mathrm{x}, \mathrm{y})$.
Since $|x|+|y|=1 \quad$ (given)
For $\quad \mathrm{x}>0, \mathrm{y}>0 ; \quad$ (given)
We have :
$$
x+y=1 \Rightarrow x+y-1=0
$$
For $x < 0, y>0$;
We have :
$$
-x+y=1 \Rightarrow x-y+1=0
$$
For $\mathrm{x} < 0, \mathrm{y} < 0$;
We have :
$$
-x-y=1 \Rightarrow x+y+1=0
$$
For $x>0, y < 0$;
We have :
$$
\mathrm{x}-\mathrm{y}=1 \Rightarrow \mathrm{x}-\mathrm{y}-1=0
$$
These four equations gives. The required locus of the point.
Since $|x|+|y|=1 \quad$ (given)
For $\quad \mathrm{x}>0, \mathrm{y}>0 ; \quad$ (given)
We have :
$$
x+y=1 \Rightarrow x+y-1=0
$$
For $x < 0, y>0$;
We have :
$$
-x+y=1 \Rightarrow x-y+1=0
$$
For $\mathrm{x} < 0, \mathrm{y} < 0$;
We have :
$$
-x-y=1 \Rightarrow x+y+1=0
$$
For $x>0, y < 0$;
We have :
$$
\mathrm{x}-\mathrm{y}=1 \Rightarrow \mathrm{x}-\mathrm{y}-1=0
$$
These four equations gives. The required locus of the point.
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