# Abs(z-4+2i)=Abs(z+8-6i). In the form asqrt13, ainQQ, what is the least value of absz?

## I think the cartesian form of the locus of $z$ is $y = \frac{3}{2} x + 5$, but I'm not sure. Thanks

Feb 16, 2018

$a = \frac{10}{13}$

#### Explanation:

$z$ is a point that is equidistant from $\left(4 - 2 i\right)$ and $\left(- 8 + 6 i\right)$ (since $| z - {z}_{0} |$ is the distance between $z$ and ${z}_{0}$)- that means in the equivalent ${\mathbb{R}}^{2}$ plane it is the perpendicular bisector of the points $\left(4 , - 2\right)$ and $\left(- 8 , 6\right)$. The answer we are seeking is the shortest distance from the origin to this line.

Now the line satisfies the equation

${\left(x - 4\right)}^{2} + {\left(y + 2\right)}^{2} = {\left(x + 8\right)}^{2} + {\left(y - 6\right)}^{2}$

which simplifies to the equation

$3 x - 2 y - 10 = 0$

The distance of this line from the origin is

$\frac{| 3 \times 0 - 2 \times 0 - 10 |}{\sqrt{{3}^{2} + {2}^{2}}} = \frac{10}{\sqrt{13}} = \frac{10}{13} \sqrt{13}$