How does this vector addition work in geometry? I saw the accepted answer to the question: Finding a point along a line a certain distance away from another point!
I am not getting how to use it actually to find the coordinates of the new point at a given distance. This is because I am confused between how to translate to/from the Cartesian system and the vector system.
So please explain me the following by walking through the solution suggested in that answer with the following example data.
Suppose I have two points $(0,0)$ and $(1,1)$ and I want to find a point at a distance which is 3/5th of the total distance between the points (i.e. $\frac{3}{5}\sqrt{2})$ from the point $(0,0)$ and lies on the segment.
How do I use the vectors mentioned in the solution given there to find the required coordinates?
Edit:
Precisely speaking, What I do expect is the explanation of:


*

*What is vector $\mathbf v$ there if $(x_1,y_1) = (1,1)$ and $(x_0,y_0) = (0,0)$

*What is the normalized vector $d\mathbf u$?

*How do I do the addition $(x_0,y_0) + d\mathbf u$?
 A: We have $\mathbf v = (x_1,y_1)-(x_0,y_0) = (1,1)-(0,0)=(1,1)$. Note that its length (or norm) is:
$$
||\mathbf v|| = \sqrt{v_1^2+v_2^2} = \sqrt{1^2+1^2}=\sqrt{2}
$$
Thus, normalizing the vector yields:
$$
\mathbf u = \frac{\mathbf v}{||\mathbf v||} = \frac{1}{\sqrt{2}}(1,1)
$$
Since we want the point that is at a distance that is three-fifths the total distance between the two points, we have $d=\dfrac{3}{5}||\mathbf v||=\dfrac{3\sqrt{2}}{5}$. Hence, the desired point is:
$$ \begin{align*}
(x_0,y_0)+d\mathbf u &= (0,0)+\dfrac{3\sqrt{2}}{5}\left(\frac{1}{\sqrt{2}}(1,1)\right)\\
&= (0,0)+\dfrac{3}{5}(1,1) \\
&= (0,0)+(3/5,3/5)\\
&=\left(\dfrac{3}{5},\dfrac{3}{5}\right)
\end{align*} $$
A: All points between $(0,0)$ and $(1,1)$ are described as a convex combination of them. That is:
$$ (x, y) = (0,0)\alpha + (1,1)(1-\alpha)$$
where $0 \leq \alpha \leq 1$. If $\alpha = 0$, then $(x,y) = (1,1)$, while if $\alpha = 1$, then $(x,y) = (0,0)$.
We can rewrite the formula as follows:
$$(x,y) = (1,1)(1-\alpha)$$
You want to find a point $(x,y)$ such that
$$ \| (x,y) - (0,0) \| = \frac{3 \sqrt{2}}{5} $$
So...
$$ \| (x,y) - (0,0) \| = \| (x,y) \| = \| (1,1)(1-\alpha) \| = \sqrt{(1-\alpha)^2 + (1-\alpha)^2} = $$
$$ = \sqrt{2(1-\alpha)^2} = \frac{3 \sqrt{2}}{5} $$
Squaring both term, we get
$$ 2(1-\alpha)^2 = \frac{9 \cdot 2}{25} $$
$$ 25(1-\alpha)^2 = 9 $$
$$ 25\alpha^2 -50 \alpha + 16 = 0 $$
Solutions are $\alpha = \frac{2}{5}$ and $\alpha = \frac{8}{5}$. Since $0 \leq \alpha \leq 1$, then $\alpha = \frac{2}{5}$  and hence
$$(x,y) = (1,1)(1 - \frac{2}{5}) = (\frac{3}{5}, \frac{3}{5})$$
is the point you are looking for.

$u$ is the normalized vector obtained as $u = (1,1) - (0,0)$, while $d$ is a distance. However, my solution does not need this vector.
