Given A(-5,-4) and B(7, 18), find the point two-fifths of the way from A to B.
+26367 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)
+128481 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
