If z_1=2-3i and z_2=-1+i,then the locus of a | vedclass

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(D) Given $z_1=2-3i$ and $z_2=-1+i$. The condition $\arg \left(\frac{z-z_1}{z-z_2}\right)=\frac{\pi}{2}$ implies that the vector $z-z_1$ is rotated by

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$x^2+y^2-x+2y-5=0$ and $4x+3y+1 > 0$

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(D) Given $z_1=2-3i$ and $z_2=-1+i$. The condition $\arg \left(\frac{z-z_1}{z-z_2}\right)=\frac{\pi}{2}$ implies that the vector $z-z_1$ is rotated by $\frac{\pi}{2}$ counter-clockwise relative to $z-z_2$. This means the angle $\angle z_1 z z_2 = \frac{\pi}{2}$.Thus,the locus of $z$ is a circle with diameter $z_1 z_2$,excluding the points $z_1$ and $z_2$ themselves.The equation of a circle with diameter endpoints $(x_1, y_1)$ and $(x_2, y_2)$ is $(x-x_1)(x-x_2) + (y-y_1)(y-y_2) = 0$.Here,$z_1 = (2, -3)$ and $z_2 = (-1, 1)$.$(x-2)(x+1) + (y+3)(y-1) = 0$$x^2 - x - 2 + y^2 + 2y - 3 = 0$$x^2 + y^2 - x + 2y - 5 = 0$.For the argument to be exactly $\frac{\pi}{2}$,the points $z, z_1, z_2$ must form a triangle in counter-clockwise order. This restricts the locus to the semi-circle lying on one side of the line passing through $z_1$ and $z_2$.The line passing through $z_1(2, -3)$ and $z_2(-1, 1)$ is $y - 1 = \frac{-3-1}{2-(-1)}(x+1)$ $\Rightarrow y-1 = -\frac{4}{3}(x+1)$ $\Rightarrow 3y-3 = -4x-4$ $\Rightarrow 4x+3y+1=0$.Testing a point like $(0,0)$ which is inside the circle: $4(0)+3(0)+1 = 1 > 0$. Thus,the condition is $4x+3y+1 > 0$.

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

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