finding some point in a line segment with given ratio

Question

Let $\sigma$ be the line segment joining the complex numbers $z_1$ and $z_2$. I want to find the point $z$ which divides $\sigma$ in the ratio $\lambda_1:\lambda_2$.

My method: If $z_1=x_1+iy_1$, $z_2=x_2+iy_2$ and $z=x+iy$ then we can write

$$(x-x_1)^2+(y-y_1)^2=\lambda_1^2,\\ (x_2-x)^2+(y_2-y)^2=\lambda_2^2$$

After some manipulation we arrive to

$$x_1^2+y_1^2-x_2^2-y_2^2+(2x_2-2x_1)x+(2y_2-2y_1)y=\lambda_1^2-\lambda_2^2$$

Intersecting this with the line

$$y-y_2=\frac{y_2-y_1}{x_2-x_1}(x-x_2)$$

we can find the point $z=x+iy$.

My question is what other methods can we use which are easier or even complex analysis methods? Thanks.

Answer 1

If you already know that the point is on $\sigma$ then there exists $t \in (0,1)$ such that: $$z=tz_1+(1-t)z_2$$ and you want: $$\frac{|z_1-z|}{|z_2-z|}=\frac{\lambda_1}{\lambda_2}$$ So: $$\frac{|z_1-(tz_1+(1-t)z_2)|}{|z_2-(tz_1+(1-t)z_2)|}=\frac{|(1-t)z_1+(1-t)z_2|}{|tz_1+t z_2|}=\frac{t}{1-t}=\frac{\lambda_1}{\lambda_2}$$ and the solution of $\frac{t}{1-t}=\frac{\lambda_1}{\lambda_2}$ i.e $\lambda_2 t-\lambda_1 (1-t)=0$ is: $$t=\frac{\lambda_1}{\lambda_1+\lambda_2}$$ so: $$z=\frac{\lambda_1}{\lambda_1+\lambda_2}z_1+\frac{\lambda_2}{\lambda_1+\lambda_2}z_2$$

Answer 2

If i understood you correctly you are looking for something like this:

$$z_{\lambda_1 : \lambda_2} = \frac{\lambda_1}{\lambda_1 + \lambda_2}z_1 + \frac{\lambda_2}{\lambda_1 + \lambda_2}z_2$$

This is a convex combination of $z_1$ and $z_2$ dividing the segment connecting these two points in the given ratio.