A point moves in x-y plane such that the sum of its distances from two mutually perpendicular lines is always equal to 2^(1/2) units. The area encloced by the locus of the point is (in sq. units)

Question

bharat bajaj · Answer

Say the mutually perpendicular lines are x and y axis :
Hence, the locus of point is : x+y = v2
Hence, it is a line.
Area enclosed by the line and both axes :
1/2 . v2. v2. = 1 unit sq.
Thanks
Bharat Bajaj
IIT Delhi

askiitians faculty

Daksh Kumar · Answer

The Lines Will Be X+y=√2 , x-y=√2-X-y=√2 , y-x=√2 and They line no. 1&4 will make a intercept of √2 units on X-axis hence length of diagonal of square will be 2√2 And Hence Side Will be 2 units So area of square= Sidexside Hence Ans. = 2x2 = 4

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A point moves in x-y plane such that the sum of its distances from two mutually perpendicular lines is always equal to 2^(1/2) units. The area encloced by the locus of the point is (in sq. units)

A point moves in x-y plane such that the sum of its distances from two mutually perpendicular lines is always equal to 2^(1/2) units. The area encloced by the locus of the point is (in sq. units)

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A point moves in x-y plane such that the sum of its distances from two mutually perpendicular lines is always equal to 2^(1/2) units. The area encloced by the locus of the point is (in sq. units)

A point moves in x-y plane such that the sum of its distances from two mutually perpendicular lines is always equal to 2^(1/2) units. The area encloced by the locus of the point is (in sq. units)

2 Answers

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Other Related Questions on Algebra

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