The extremities of a diagonal of a rectangle are (0.0) and (4, 4). The locus of the extremities of the other diagonal is equal to
A
`x^(2) +y^(2) - 4x - 4y = 0`
B
`x^(2)+y^(2)+4x +4y - 4 = 0`
C
`x^(2)+y^(2)+4x +4y +4 = 0`
D
`x^(2)+y^(2) - 4x - 4y - 4 = 0`
Text Solution
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The correct Answer is:
A
Rectangle ABCD has vertices `A(0,0)` and `D(4,4)`. Let B is `(h,k)` `AB _|_ BC` `:. (k)/(h).(k-4)/(h-4) =-1` `:.`The locus of the exterminates of the other diagonal is `(x-0) (x-4) +(y-4)(y-0) =0` or `x^(2)+y^(2) -4x -4y = 0`
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