Given A(-5,-4) and B(7, 18), find the point two-fifths of the way from A to B.

Given A(-5,-4) and B(7, 18), find the point two-fifths of the way from A to B.

$\begin{array}{|rcll|} \hline \vec{A} &=& \binom{-5}{-4} \\ \vec{B} &=& \binom{7}{18} \\ \hline \end{array}$

$\begin{array}{|rcll|} \hline \vec{x} &=& \vec{A}+(\vec{B}-\vec{A})\cdot \lambda \qquad & | \qquad 0\le \lambda\le 1 \\ && & | \qquad \lambda = 0 \Rightarrow \vec{x} = \vec{A}\\ && & | \qquad \lambda = 1 \Rightarrow \vec{x} = \vec{B}\\ \vec{x} &=& \vec{A} + \lambda\cdot \vec{B} + \lambda\cdot\vec{A} \\ \vec{x} &=& (1-\lambda)\cdot\vec{A}+ \lambda\cdot \vec{B} \qquad & | \qquad \lambda = \dfrac25 \\\\ \vec{x} &=& (1-\frac25) \cdot\vec{A}+ \frac25 \cdot \vec{B} \\ \vec{x} &=& \frac35 \cdot\vec{A} + \frac25 \cdot \vec{B} \\ \vec{x} &=& 0.6 \cdot\vec{A} + 0.4 \cdot \vec{B} \\ \vec{x} &=& 0.6 \cdot\binom{-5}{-4} + 0.4 \cdot \binom{7}{18} \\ \vec{x} &=& \binom{-3}{-2.4} + \binom{2.8}{7.2} \\ \vec{x} &=& \binom{-3+2.8}{-2.4+7.2} \\\\ \mathbf{\vec{x}} &\mathbf{=}& \mathbf{\dbinom{-0.2}{4.8} }\\ \hline \end{array}$

The Point is (-0.2, 4.8)

Given A(-5,-4) and B(7, 18), find the point two-fifths of the way from A to B.

This can be found thusly :

The x coordinate is given by ;

-5 + [2/5] [ 7 - (-5)]  =   -5 + [2/5] [12]  =  -1/5  = -.2

And the y coordinate is given by :

-4 + [2/5] [ 18 - (-4) ]  =  -4 + [2/5] [22]  = 4.8  

Proof :

The total distance between the points =  2√157

And the distance between  (-5, 4)   and ( -.2, 4.8)  = √  [4.8^2 + .8^2]  ≈ 10.024

So.....10.024 / [2 √157] ≈  .4    ≈   2/5