Question

If $z_1=2-3i$ and $z_2=-1+i$, then the locus of a point $P$ represented by $z=x+iy$ in the argand plane satisfying the equation $\arg \left(\frac{z-z_1}{z-z_2}\right)=\frac{\pi }{2}$ is

• A. $x^2+y^2-x+2y-5=0$
• B. $x^2+y^2-x+2y-5=0$ and $4x+3y+1<0$
• C. $4x+3y+1=0$ and $x^2+y^2-x+2y-5>0$
• D. $x^2+y^2-x+2y-5=0$ and $4x+3y+1>0$

Answer 1

Answer: (D)

We have,
$z_1=2-3i,z_2=-1+i$
and $\arg \left(\frac{z-z_1}{z-z_2}\right)=\frac{\pi }{2}$
$\therefore {\left|z-z_1\right|}^2+{\left|z_2-z_2\right|}^2={\left|z_1-z_2\right|}^2$
$\Rightarrow |z-(2-3i){|}^2+|z-(-1+i){|}^2=|2-3i+1-i{|}^2$
$\Rightarrow (x-2{)}^2+(y+3{)}^2+(x+1{)}^2+(y-1{)}^2=9+16$
$[\because z=x+iy]$
$\Rightarrow x^2-4x+4+y^2+6y+9+x^2+2x+1+y^2$
$-2y+1=25$
$\Rightarrow 2x^2+2y^2-2x+4y-10=0$
$\Rightarrow x^2+y^2-x+2y-5=0$
Now equation of line paring through
$z_1(2,-3)$ and $z_2(-1,1)$ is
$y+3=\frac{1+3}{-1-2}(x-2)$
$\Rightarrow y+3=\frac{4}{-3}(x-2)$
$\Rightarrow -3y-a=4x-8$
$\Rightarrow -3y-9=4x-8$
$\Rightarrow 4x+3y+1=0$
According to given condition, $z$ should not lies on $z_1$ and $z_2$
$\therefore 4x+3y+1>0$