Question
Question: Show that the perpendicular distance of a point d (d is a complex number) on the Argand plane from t...
Show that the perpendicular distance of a point d (d is a complex number) on the Argand plane from the line az+az+b=0 (a is a complex number and b is a real) is 2∣a∣ad+ad+b .
Solution
Hint : In order to solve the question given above, you need to know about complex numbers. They refer to numbers that can be expressed in the form x+iy , where x and y are real numbers and i is an imaginary unit. You also need to understand the concept of complex plane or argand plane, which refers to a plot of complex numbers as points.
Complete step by step solution:
We know that the equation of the line is: az+az+b=0 .
Now, we have to put z=x+iy .
The above equation becomes:
a(x−iy)+a(x+iy)+b=0 .
Therefore, from the above equation, we get that:
(a+a)x+(a−a)iy+b=0 .
Now, let d=d1+id2 in the equation,
With the help of the above information, you can calculate the distance of the line from d:
(a+a)2+[(a−a)i]2∣(a+a)d1+(a−a)id2+b∣ .
This can be written as:
(a+a)2−(a−a)2∣a(d1−id2)+a(d1+id2)+b∣
⇒4aaad+ad+b .
This gives us:
4a∣2ad+ad+b
⇒2∣a∣ad+ad+b , which is our required answer.
Hence, proved.
Note : To solve questions similar to the one given above, you need to have basic understanding about a few important topics. These topics are: 1) complex numbers: they are numbers which include both the real and the imaginary parts. They can be expressed in the form x+iy , where x and y are real numbers and i is an imaginary unit and 2) argand plane: it refers to a diagram which contains the plot of complex numbers as points.