If the amplitude of z-1-2i is (π)/3, then the locus of z is

Correct answer: (A) y=√3x+(2-√3). Amplitude of (z-1-2i) is (π)/3 (x+iy-2i-1) =(x-1)+i(y-2) Now tan(π)/3=(y-2)/(x-1) ⇒√3=±(y-2)/(x-1) ⇒√3x-√3=y-2 √3x+(2-√3)=y …i and √3=-(y-2)/(x-1) √3x-√3=-y+2 y+√3x-2+√3=0 …ii

Q46 / 160 · TS EAMCET 2019 (03 May Shift 2) · Mathematics

If the amplitude of $(z - 1 - 2 i)$ is $\frac{π}{3}$ , then the locus of $z$ is

Correct answer
A
$y = \sqrt{3} x + (2 - \sqrt{3})$
B
$y = \sqrt{3} x - \sqrt{3}$
C
$x = \sqrt{3} y + (2 - \sqrt{3})$
D
$y = \sqrt{3} x + 2$

Solution

Amplitude of $(z - 1 - 2 i)$ is $\frac{π}{3}$

$(x + i y - 2 i - 1)$

$= (x - 1) + i (y - 2)$

Now $\tan \frac{π}{3} = |\frac{y - 2}{x - 1}|$

$\Rightarrow \sqrt{3} = \pm \frac{y - 2}{x - 1}$

$\Rightarrow \sqrt{3} x - \sqrt{3} = y - 2$

$\sqrt{3} x + (2 - \sqrt{3}) = y \dots (i)$

and $\sqrt{3} = - \frac{y - 2}{x - 1}$

$\sqrt{3} x - \sqrt{3} = - y + 2$

$y + \sqrt{3} x - (2 + \sqrt{3}) = 0 \dots (ii)$

Hence, the correct answer is option: 'A'.

If the amplitude of z-1-2i is π3, then the locus of z is

Solution

Relevant Questions

If the amplitude of $(z - 1 - 2 i)$ is $\frac{π}{3}$ , then the locus of $z$ is