The locus of a point so that sum of its dista | vedclass

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Question

(b) We take the coordinate axes as two perpendicular lines. Let $P\,({x_1},{y_1})$ be the required point.

From $P\,({x_1},{y_1})$, we draw $PM$ and

Vedclass · Accepted Answer

$x + y = 2$

Vedclass · Answer

(b) We take the coordinate axes as two perpendicular lines. Let $P\,({x_1},{y_1})$ be the required point.

From $P\,({x_1},{y_1})$, we draw $PM$ and $PN$ perpendicular to $OX$ and $OY$ respectively.

Given, $PM + PN = 2$ ......$(i)$

But, $PM = {y_1},PN = {x_1}$

Hence from $(i)$, ${y_1} + {x_1} = 2$

Thus locus of $({x_1},{y_1})$is $x + y = 2$

which is a straight line.

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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