If the sum of the distances of a point from t | vedclass

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(a) 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| +

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Square

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(a) 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 $|x| + |y| = 1$ ==> $ - 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 $ABCD.$

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

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