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Q. One vertex of the triangle of maximum area that can be inscribed in the curve...

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user8235304 | Student, Grade 11 | Valedictorian

Posted August 22, 2013 at 5:06 PM via web

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Q. One vertex of the triangle of maximum area that can be inscribed in the curve `|z-2i|=2;` is `2+2i` ,remainning vertices is/are:-

A) `-1 + i(2+sqrt3)`

B) `-1-i(2+sqrt3)`

C) `1+i(2+sqrt3)`

D) `-1-i(2-sqrt3)`

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sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted August 22, 2013 at 5:46 PM (Answer #1)

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You need to come up with the following notation for the two remaining vertices, `a + bi` and `c + di.`

These vertices need to lie on the given curve, such that:

`{(|a + bi - 2i| = 2),(|c + di - 2i| = 2):} => |a + bi - 2i| = |c + di - 2i|`

Evaluating the absolute value of the complex number `a + i*(b - 2) ` yields:

`|a + bi - 2i| = sqrt(a^2 + (b - 2)^2)`

Since `|a + bi - 2i| = 2` , yields:

`sqrt(a^2 + (b - 2)^2) = 2`

Replacing `1` for a and `(2 + sqrt3)` for b yields:

`|1 + (2 +sqrt3)i - 2i| = 2`

`sqrt(1^2 + (2 + sqrt3 - 2)^2)) = 2`

`sqrt(1 + 3) = 2 => sqrt 4 = 2 => 2 = 2`

Replacing -`1` for a and `(2 + sqrt3)` for b yields:

`|-1 + (2 + sqrt3)i - 2i| = 2`

`sqrt((-1)^2 + (2 + sqrt3 - 2)^2)) = 2 => 2 = 2`

Replacing `-1` for a and `(-2 - sqrt3)` for b yields:

`sqrt((-1)^2 + (-2 - sqrt3 - 2)^2)) = 2`

`sqrt(1+ (4 + sqrt3)^2)) != 2 =>` the vertex `z = -1 - i*(2 + sqrt3) ` does not beong to the curve

Replacing -`1` for a and `(-2 + sqrt3)` for b yields:

`sqrt((-1)^2 + (-2 + sqrt3 - 2)^2)) = 2`

`sqrt(1+ (sqrt3 - 4)^2)) != 2 => the vertex z = -1 - i*(2 - sqrt3) ` does not beong to the curve

Hence, evaluating the other two possible vertices that belong to the curve yields `z_(1,2) = +-1 + (2 + sqrt3)*i.`

Since the area is maximum, the only valid vertex is `z = 1 + i*(2 + sqrt 3),` hence, you need to select the answer `C)` .

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