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The locus of a point so that sum of its distance from two given perpendicular lines is equal to 2 unit in first quadrant, is
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Verified Answer
The correct answer is:
$x+y=2$
We take the coordinate axes as two perpendicular lines. Let $P\left(x_1, y_1\right)$ be the required point. From $P\left(x_1, y_1\right)$, we draw $P M$ and $P N$ perpendicular to $O X$ and $O Y$ respectively.
Given, $P M+P N=2$
But, $\quad P M=y_1, P N=x_1$
Hence from (i), $y_1+x_1=2$
Thus locus of $\left(x_1, y_1\right)$ is $x+y=2$ which is a straight line.
Given, $P M+P N=2$
But, $\quad P M=y_1, P N=x_1$
Hence from (i), $y_1+x_1=2$
Thus locus of $\left(x_1, y_1\right)$ is $x+y=2$ which is a straight line.
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