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Question: \(\mathbf{z}_{\mathbf{1}}\mathbf{,}\mathbf{z}_{\mathbf{2}}\mathbf{,}\) are the inverse points with r...

z1,z2,\mathbf{z}_{\mathbf{1}}\mathbf{,}\mathbf{z}_{\mathbf{2}}\mathbf{,} are the inverse points with respect to the line

zaˉ+azˉ=bz\bar{a} + a\bar{z} = bif

A

z1a+z2a=bz_{1}a + z_{2}a = b

B

z1aˉ+az2‾=bz_{1}\bar{a} + a\overline{z_{2}} = b

C

z1aˉ−az2‾=bz_{1}\bar{a} - a\overline{z_{2}} = b

D

None of these

Answer

z1aˉ+az2‾=bz_{1}\bar{a} + a\overline{z_{2}} = b

Explanation

Solution

Sol. Let RS be the line represented by the equation zaˉ+azˉ=bz\bar{a} + a\bar{z} = b

Let P and Q are the inverse points with respect to the line RS.

The point Q is the reflection (inverse) of the point P in the line RS if the line RS is the right bisector of PQ. Take any point z in the line RS, then lines joining z to P and z to Q are equal.

i.e., ∣z−z1∣=∣z−z2∣|z - z_{1}| = |z - z_{2}| or ∣z−z1∣2=∣z−z2∣2|z - z_{1}|^{2} = |z - z_{2}|^{2}

i.e.,(z−z1)(zˉ−z1‾)=(z−z2)(zˉ−z2‾)(z - z_{1})(\bar{z} - \overline{z_{1}}) = (z - z_{2})(\bar{z} - \overline{z_{2}}) ⇒ z(z2‾−z1‾)+zˉ(z2−z1)+(z1z1‾−z2z2‾)=0z(\overline{z_{2}} - \overline{z_{1}}) + \bar{z}(z_{2} - z_{1}) + (z_{1}\overline{z_{1}} - z_{2}\overline{z_{2}}) = 0 .....(ii)

Hence, equations (i) and (ii) are identical, therefore

comparing coefficients, we get

aˉz2‾−z1‾=az2−z1=−bz1z1‾−z2z2‾\frac{\bar{a}}{\overline{z_{2}} - \overline{z_{1}}} = \frac{a}{z_{2} - z_{1}} = \frac{- b}{z_{1}\overline{z_{1}} - z_{2}\overline{z_{2}}} So that,

z1aˉz1(z2‾−z1‾)=az2‾z2‾(z2−z1)=−bz1z1‾−z2z2‾=z1aˉ+az2‾−b0\frac{z_{1}\bar{a}}{z_{1}(\overline{z_{2}} - \overline{z_{1}})} = \frac{a\overline{z_{2}}}{\overline{z_{2}}(z_{2} - z_{1})} = \frac{- b}{z_{1}\overline{z_{1}} - z_{2}\overline{z_{2}}} = \frac{z_{1}\bar{a} + a\overline{z_{2}} - b}{0}

(By ratio and proportion rule)

Hence, z1aˉ+az2‾−bz_{1}\bar{a} + a\overline{z_{2}} - b = 0 or z1aˉ+az2‾=b.z_{1}\bar{a} + a\overline{z_{2}} = b.