Assertion A- z - x - iy is such that - z -1 - z -1 - -1 then locus of z is a circleReason R- If - z - z 1 - z - z 2 - -1 then locus of z is perpendicular bisector of z1- z2

The correct option is A A is false, R is true | #160; #160;z+1/z 1 |=1 ∴ | z+1 |= | z 1 | ∴ z lies on the perpendicular bisector joining ...

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Byju's Answer

Standard XII

Mathematics

Inverse of a Function

Assertion A: ...

Question

Assertion (A): z = x + iy is such that $∣ ∣ \frac{z + 1}{z - 1} ∣ ∣ = 1$ then locus of z is a circle
Reason (R): If $∣ ∣ \frac{z - z_{1}}{z - z_{2}} ∣ ∣ = 1$ then locus of z is perpendicular bisector of $z_{1}, z_{2}$

Both A and R are true and R is the correct explanation of A

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Both A and R are true and R is not a correct explanation of A

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A is false, R is true

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A is true R is false

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Solution

The correct option is A A is false, R is true
$∣ ∣ ∣ \frac{z + 1}{z - 1} ∣ ∣ ∣ = 1$
$∴ | z + 1 | = | z - 1 |$
$∴ z$ lies on the perpendicular bisector joining $(1, 0)$ and $(- 1, 0)$ which is $y = 0$ because distance of $z$ from $(1, 0)$ is equal to distance from $(- 1, 0)$
Hence, the Assertion is false but the Reason is true.

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Similar questions

Q. Let

z_{1}

and

z_{2}

are fixed complex numbers, the locus of

z

so that

\frac{z - z_{1}}{z - z_{2}} = e^{i θ}, θ \in R

Q. Assertion (A): At constant temperature,

P V

P

plot for real gases is not a straight line.
Reason (R): At high pressure all gases have

(Z > 1)

but at intermediate pressure most gases have

(Z < 1)

In the following question, a statement of Assertion (A) followed by a statement of Reason (R) is given. Choose the correct option out of the choices given below in each question.

Assertion (A) At constant temperature, PV vs V plot for real gases is not a straight line.

Reason (R) At high pressure all gases have Z > 1 but at an intermediate pressure, most gases have Z < 1.

(a) Both A and R are true and R is the correct explanation of A

(b) Both A and R are true R is not the correct explanation of A

(d) A is false but R is true

Q. If

| z - z_{1} | + | z - z_{2} | = k (> | z_{1} - z_{2} |)

, then the locus of

z

is a/an:

(Given:

z_{1}

and

z_{2}

are fixed complex numbers)

Q. The maximum area of the triangle formed by the complex coordinates

z, z_{1}, z_{2}

which satisfy the relations

| z - z_{1} | = | z - z_{2} |

and

| z - (z_{1} + z_{2}) / 2 | \leq r

where

r > | z_{1} - z_{2} |

, is

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Assertion (A): z = x + iy is such that ∣∣z+1z−1∣∣=1 then locus of z is a circleReason (R): If ∣∣z−z1z−z2∣∣=1 then locus of z is perpendicular bisector of z1,z2

The correct option is A A is false, R is true∣∣∣z+1z−1∣∣∣=1∴|z+1|=|z−1|∴z lies on the perpendicular bisector joining(1,0) and (−1,0) which is y=0 because distance of z from (1,0) is equal to distance from (−1,0)Hence, the Assertion is false but the Reason is true.

Assertion (A): z = x + iy is such that $∣ ∣ \frac{z + 1}{z - 1} ∣ ∣ = 1$ then locus of z is a circle
Reason (R): If $∣ ∣ \frac{z - z_{1}}{z - z_{2}} ∣ ∣ = 1$ then locus of z is perpendicular bisector of $z_{1}, z_{2}$