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I know that

The equation of a line joining z$_1$ & z$_2$ is given by, $z = z_1 + t (z_ 1 - z_2)$ where t is a parameter

Then how can we wrte a line through the point z1 & perpendicular to the line joining z1 to the origin.

Also how can we it in form of determinant of line joining two points

2 Answers 2

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You can easily think of this is terms of vectors(i find that intuitive). The vector eqn. Of a line is given by:

$\vec r = \vec a + \lambda( \vec b)$ where $\vec a$ is a given point through which the line passes and $\vec b$ is a vector to which the line is parallel. Now just imagine complex nos. In place of vectors, and you can easily find the eqn. Of line through$z_1$ and $z_2$. To be : $z = z_1 + \lambda(z_1-z_2)$

To find the eqn. Of the line passing through $z_1$ and perpendicular to its position vector(parallel to $iz_1$ ) we write:

$z = z_1 + \lambda(i.z_1)$ this is the required eqn.

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    How do you wrote position vector as $iz_1$ – 2017-01-23
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    Correspond it to the first eqn. I gave. The eqn of a line r,isgiven by 2parameters.a point through which it passes (a) and a vector to which it is parallel. In the complex case, the line you askfor passes through z_1 and is perpendicular to the line joining z1 to the origin(its position vector). Clearly this means that the line is parallel to iz_1 which isperpendicular to z_1. Hence the last eqn. – 2017-01-23
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The line joining the origin to $z_1$ also passes through the point $2z_1$. Hence the problem can be rephrased in this manner : find the equation of the perpendicular bi sector of the line joining the origin and the point $2z_1$. Hence we get :

$$ \lvert z \rvert = \lvert z-2z_1 \rvert $$ Can you take it from here?

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    Now how can we write it in that form – 2017-01-23