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Section Formula Using Complex Numbers

In ABC, Az1...

Question

In $△ A B C, A (z_{1}), B (z_{2}),$ and $C (z_{3})$ are inscribed in the circle $| z | = 5$ . If $H (z_{H})$ be the orthocenter of triangle ABC, then find $z_{H}$ .

A

z_{H} = - (z_{1} + z_{2} + z_{3})

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B

z_{H} = 0

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C

z_{H} = z_{1} + z_{2} + z_{3}

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D

z_{H} = 2 (z_{1} + z_{2} + z_{3})

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Solution

The correct option is B $z_{H} = z_{1} + z_{2} + z_{3}$
Circumcenter of $△ A B C$ is clearly origin. Let, $G (z_{G})$ be its centroid. Then,
$z_{G} = \frac{1}{3} (z_{1} + z_{2} + z_{3})$
Now we know that
OG : GH = 1:2
$⟹ z_{G} = \frac{2 \times 0 + 1 \times z_{H}}{3}$
$⟹ z_{H} = 3 z_{G} = z_{1} + z_{2} + z_{3}$
Ans: C

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Similar questions

△ A B C, A (z_{1}), B (z_{2})

C (z_{3})

are inscribed in the circle

| z | = 5

H (z_{H})

be the orthocentre of

△ A B C

z_{H} =

△ A B C, A (z_{1}), B (z_{2})

C (z_{3})

are inscribed in the circle

| z | = 5

H (z_{H})

be the orthocentre of

△ A B C

z_{H} =

Q. Statement 1: If

| z_{1} | = | z_{2} | = | z_{3} |, z_{1} + z_{2} + z_{3} = 0

(z_{1}), B (z_{2}), C (z_{3})

are the vertices of

Δ A B C

, then one of the values of

a r g (\frac{(z_{2} + z_{3} - 2 z_{1})}{(z_{3} - z_{2})})

\frac{π}{2}

.
Statement 2: In equilateral triangle, orthocenter coincides with centroid.

A (z_{1}), B (z_{2}), C (z_{3})

are the vertices of the triangle

A B C

(in anticlockwise order). If

∠ A B C = \frac{π}{4}

A B = \sqrt{2} (B C)

A (z_{1}), B (z_{2}), C (z_{3})

are the vertices of the triangle ABC such that

\frac{(z_{1} - z_{2})}{(z_{3} - z_{2})} = (\frac{1}{\sqrt{2}} + \frac{i}{\sqrt{2}})

, then triangle

A B C

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Section Formula Using Complex Numbers

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