Question

Show that the perpendicular distance of a point d (d is a complex number) on the Argand plane from the line $\bar az + a\bar z + b = 0$ (a is a complex number and b is real) is $\dfrac{{a\bar d + \bar ad + b}}{{2\left| a \right|}}.$

Answer 1

Hint : In this question write the point d in the complex form and then find the distance of the point from the given straight line by using the distance formula of the point from the straight line and after certain simplification we can arrive at the result.

Complete step-by-step answer :
Given
The equation of the line is $\bar az + a\bar z + b = 0 - - (i)$
Now we know complex number $z = x + iy$and its conjugate is given as $\bar z = x - iy$
Now substitute the complex number in the line equation (i), we get
$\bar a\left( {x + iy} \right) + a\left( {x - iy} \right) + b = 0$
Hence by further solving this equation we get
$\left( {\bar a + a} \right)x + \left( {\bar a - a} \right)iy + b = 0$
Now the point d which is a complex number can be written as
$d = {d_1} + i{d_2}$
So the distance of the line from the point d can be written as
$\Rightarrow d = \dfrac{{\left| {\left( {\bar a + a} \right){d_1} + \left( {\bar a - a} \right)i{d_2} + b} \right|}}{{\sqrt {{{\left( {\bar a + a} \right)}^2} + {{\left[ {\left( {\bar a - a} \right)i} \right]}^2}} }}$
Now solve this equation further, we get
$\Rightarrow d = \dfrac{{\left| {a\left( {{d_1} - i{d_2}} \right) + \bar a\left( {{d_1} + i{d_2}} \right) + b} \right|}}{{\sqrt {{{\left( {\bar a + a} \right)}^2} - {{\left( {\bar a - a} \right)}^2}} }}$
Now since $d = {d_1} + i{d_2}$, so we can write $\bar d = {d_1} - i{d_2}$, by substituting this we can further write

$$\Rightarrow d = \dfrac{{\left| {a\bar d + \bar ad + b} \right|}}{{\sqrt {{{\left( {\bar a} \right)}^2} + {a^2} + 2\bar aa - {{\left( {\bar a} \right)}^2} - {a^2} + 2\bar aa} }} \\\ \Rightarrow d = \dfrac{{\left| {a\bar d + \bar ad + b} \right|}}{{\sqrt {4\bar aa} }} \\\$$

Now since${\left| a \right|^2} = a\bar a$, hence we can write the distance as

$$\Rightarrow d = \dfrac{{\left| {a\bar d + \bar ad + b} \right|}}{{\sqrt {4{{\left| a \right|}^2}} }} \\\ = \dfrac{{\left| {a\bar d + \bar ad + b} \right|}}{{2\left| a \right|}} \;$$

So the perpendicular distance of a point d on the Argand plane from the line $\bar az + a\bar z + b = 0$is $= \dfrac{{\left| {a\bar d + \bar ad + b} \right|}}{{2\left| a \right|}}$
Hence proved

Note : Shortest distance from a point to a perpendicular line is given by the formula $d = \dfrac{{\left| {A{x_1} + B{y_1} + C} \right|}}{{\sqrt {{A^2} + {B^2}} }}$ where, x and y indicates the coordinates of the points through which the line is passing.