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Thread: Complex no.

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    Complex no.

    If $\displaystyle Z_{1},Z_{2}$ and $\displaystyle Z_{3}$ are Complex no. Representing The Vertices of a Triangle inscribed in $\displaystyle |Z|=2$. The

    Altitude Through $\displaystyle Z_{1}$ meets the Circumcircle in $\displaystyle P$, Then The Complex no. Corrosponding To $\displaystyle P$ is =

    (Ans in Terms of $\displaystyle Z_{1},Z_{2}$ and $\displaystyle Z_{3}$)
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    Quote Originally Posted by jacks View Post
    If $\displaystyle Z_{1},Z_{2}$ and $\displaystyle Z_{3}$ are Complex no. Representing The Vertices of a Triangle inscribed in $\displaystyle |Z|=2$. The

    Altitude Through $\displaystyle Z_{1}$ meets the Circumcircle in $\displaystyle P$, Then The Complex no. Corrosponding To $\displaystyle P$ is =

    (Ans in Terms of $\displaystyle Z_{1},Z_{2}$ and $\displaystyle Z_{3}$)
    Dear jacks,

    Let, $\displaystyle z_1=x_1+iy_1$

    $\displaystyle z_2=x_2+iy_2$

    $\displaystyle z_3=x_3+iy_3$

    Then write the gradient of the $\displaystyle Z_{2}Z_{3}$ line. Let it be $\displaystyle m_1$.

    Since $\displaystyle Z_{2}Z_{3}$ and $\displaystyle PZ_{1}$ are perpendicular you could find the gradient of the $\displaystyle PZ_{1}$ line by,

    $\displaystyle m_{1}\times m=-1$

    Then you could fine the equation of the $\displaystyle PZ_{1}$ line. Hint: You know the gradient and a point on the $\displaystyle PZ_{1}$ line. Let the equation be $\displaystyle y=mx+c$----(1)

    The equation of the circle could be written as, $\displaystyle \mid Z\mid=2\Rightarrow{x^2+y^2=4}$-----------(2)

    Using equations (1) and (2) you can find the point P.
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