In the figure | z |= r is circumcircle of triangle ABC . D , E F are the middle points of the sides BC , CA AB respectively, AD produced to meet the circle at L. If angle CAD =θ, AD = x , BD = y and altitude of triangle ABC from A meet the circle |z|=r at M, za, zb zc are affixes of vertices A, B C respectively. Affix of M is -

Question

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/mathematics/b0071801428da9e28f65ed2cd94615a6-.png" /> In the figure | z |= r is circumcircle of triangle ABC . D , E F are the middle points of the sides BC , CA AB respectively, AD produced to meet the circle at L. If angle CAD =θ, AD = x , BD = y and altitude of triangle ABC from A meet the circle |z|=r at M, za, zb zc are affixes of vertices A, B C respectively. Affix of M is - (A) 2 zb e i 2 B  (B) zb ei(π-2 B) (C) zb ei B (D) 2 zb ei B

Tardigrade Admin · Accepted Answer

Let affix of M is zm and angle B O M=π-2 B, then (zm-0/zb-0)=(O M/O B) ei(π-2 B) zm=zb ei(π-2 B)

$2 z_{b} e^{i 2 B}$

$z_{b} e^{i (π - 2 B)}$

$z_{b} e^{i B}$

$2 z_{b} e^{i B}$

Let affix of $M$ is $z_{m}$ and $∠ BOM = π - 2 B$ , then
$\frac{z _{m} - 0}{z _{b} - 0} = \frac{OM}{OB} e^{i (π - 2 B)}$
$z_{m} = z_{b} e^{i (π - 2 B)}$

2zb​ei2B

zb​ei(π−2B)

zb​eiB

2zb​eiB

Let affix of M is zm​ and ∠BOM=π−2B, then zb​−0zm​−0​=OBOM​ei(π−2B) zm​=zb​ei(π−2B)

$2 z_{b} e^{i 2 B}$

$z_{b} e^{i (π - 2 B)}$

$z_{b} e^{i B}$

$2 z_{b} e^{i B}$

Let affix of $M$ is $z_{m}$ and $∠ BOM = π - 2 B$ , then
$\frac{z _{m} - 0}{z _{b} - 0} = \frac{OM}{OB} e^{i (π - 2 B)}$
$z_{m} = z_{b} e^{i (π - 2 B)}$