AP EAMCET · Maths · Complex Number
If \(a\) and \(c\) are complex numbers and \(b\) is a real number in the Argand plane, then the perpendicular distance from \(c\) to the line \(a \bar{z}+\bar{a} z+b=0\) is
- A \(\frac{(a \bar{c}+\bar{a} c+b)}{2|a|}\)
- B \(\frac{(\bar{a} \bar{c}+a c+b)}{2|a|}\)
- C \(\frac{(a \bar{c}+\bar{a} c+b)}{|a|}\)
- D \(\frac{(\bar{a}+b+\bar{c})}{2|a|}\)
Answer & Solution
Correct Answer
(A) \(\frac{(a \bar{c}+\bar{a} c+b)}{2|a|}\)
Step-by-step Solution
Detailed explanation
Distance \( = \frac{(a \bar{c} + \bar{a} c + b)}{2|a|} \)
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