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A-point-moves-in-xy-plane-such-that-sum-of-its-distance-from-two-mutually-perpendicular-lines-is-always-3-The-area-encloded-by-the-locus-of-the-point-is-




Question Number 31371 by momo last updated on 07/Mar/18
A point moves in xy plane such   that sum of its distance from two   mutually perpendicular lines is  always 3.The area encloded by  the locus of the point is
$${A}\:{point}\:{moves}\:{in}\:{xy}\:{plane}\:{such}\: \\ $$$${that}\:{sum}\:{of}\:{its}\:{distance}\:{from}\:{two}\: \\ $$$${mutually}\:{perpendicular}\:{lines}\:{is} \\ $$$${always}\:\mathrm{3}.{The}\:{area}\:{encloded}\:{by} \\ $$$${the}\:{locus}\:{of}\:{the}\:{point}\:{is} \\ $$
Answered by ajfour last updated on 07/Mar/18
Area enclosed by  ∣x∣+∣y∣=3  is = (3(√2))^2  = 18 sq.units .
$${Area}\:{enclosed}\:{by} \\ $$$$\mid{x}\mid+\mid{y}\mid=\mathrm{3}\:\:{is}\:=\:\left(\mathrm{3}\sqrt{\mathrm{2}}\right)^{\mathrm{2}} \:=\:\mathrm{18}\:{sq}.{units}\:. \\ $$
Commented by momo last updated on 07/Mar/18
explain please
$${explain}\:{please} \\ $$

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