Question

The equation $\left|\frac{z-z_1}{z-z_2}\right|=k$ represents a straight line, if

• A. $k=1$
• B. $k\ne 1$
• C. $k>1$
• D. $k<1$

Answer 1

The correct option is A $k=1$

Let us take $z=x+iy,$
$z_1=x_1+i{y}_1$ & $z_2=x_2+i{y}_2$
$|z-z_1|=k|z-z_2|$
Squaring both sides
$(x-x_1{)}^2+(y-y_1{)}^2=k^2\left[\right(x-x_2{)}^2+(y-y_2{)}^2$]
$x^2+y^2-2x{x}_1-2y{y}_1+x{{}_1}^2+y{{}_1}^2=k^2[x^2+y^2-2x{x}_2-2y{y}_2+x{{}_2}^2+y{{}_2}^2]$
$(k^2-1)x^2+(k^2-1)y^2-2x(x_2{k}^2-x_1)-2y(y_2{k}^2-y_1)+k^{2}x{{}_2}^2+k^{2}y{{}_2}^2-x{{}_1}^2-y{{}_1}^2=0$

The equation will represent a straight line if the coefficients of $x^2$ and $y^2$ are zero
$k^2-1=0$
$k=\pm 1$

However, $k$ cannot be negative as it is equal to a non-negative number in the equation.

Hence, $k=1$