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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 locus of mid-point of the line intercepted between real and imaginary axis is given byA. `az- bar(az) =0`B. `az + bar(az) =0`C. `az-bar(az) + b =0`D. `az - bar(az) + 2b = 0` |
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Answer» Correct Answer - B 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` |
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