Question

Consider the equaiton of line `abarz + abarz+ abarz + b=0`, where b is a real parameter and a is fixed non-zero complex number.
The intercept of line on imaginary axis is given by
A. `(b)/(bara-a)`
B. `(2b)/(bara-a)`
C. `(b)/(2(bara - a))`
D. `(b)/(a-bara)`

Answer 1

Correct Answer - D
Given equation of line is `abarz + abarz + b =0AA b in R`.
Let the PQ be the segement intercept between axes.
For intercept on real axis `Z_(R)`.
`z = barz`
`rArr Z_(R)(a+ bara) + b =0`
` rArr Z_(R) = (-b)/(a + bara)`
For intercept on imaginary `Z_(1)`
`z + barz = 0`
`rArr Z_(1)(bara - a) + b=0`
`rArr Z_(1) = (b)/(a+bara)`
For mid-point,
`z= (Z_(R) + Z_(I))/(2)`
`rArr z = (-b)/(2)[(1)/(bara+a)+(1)/(bara +a)]`
`z= (barab)/((a + bara)(a-bara))`
`rArr z = (barab)/(a^(2) -(a)^(2))`
`(z[a^(2)-(a)^(2)])/(bara) = barz((bara)^(2) - (a)^(2))/(a)`
`rArr az + bar(az) =0`