How to divide a line into 3 equal parts

Question

Find the points P1(x1,y1) and P2(x2,y2) on the line segment joining A(2,-1) and B(6,5) that divide the line segment into 3 equal parts.

I wasn’t able to figure out how to make this line into 3 parts. I got as far as finding the distance between the two being sqrt(52) and the midpoint being (4,2). Is there a certain way to split it into thirds instead of halves?

Answer 1

Guide:

• Find vector $\vec{AB}$.
• Then $\vec{OP_1} = \vec{OA} + \frac13\vec{AB}$
• $\vec{OP_2} = \vec{OB} - \frac13\vec{AB}=\vec{OP_1} + \frac13\vec{AB}$

Answer 2

The parametric equation of the segment joining your two points is $$ x=2+4t$$ and $$y=-1+6t$$ where the segment is traces as $t$ runs from $0$ to $1$

For $t=1/3$ you get your first point $$ P_1=(10/3, 1)$$ and for $t=2/3$ yo get yor second point $$P_2=(14/3, 3)$$

Answer 3

One way to think about this could be the concept of Center of Mass. To calculate $P_1$ assume a mass of $2\space units$ at point $\textbf{A}$ and mass of $1 \space unit$ at point $\textbf{B}$. Then we know the Center of Mass will be located at a point that is at a distance of $\displaystyle\frac{1}{3}\textbf{AB}$ from point $\textbf{A}$.

Then

$$ P_1(x,y) = P_1 \left( \frac{m_1x_1 +m_2x_2}{m_1+m_2},\frac{m_1y_1 +m_2y_2}{m_1+m_2}\right) = P_1\left(\frac{2\times2 + 1\times 6}{2+1}, \frac{2\times-1 + 1\times5}{2+1}\right) = P_1\left(\frac{10}{3},1\right) $$

Similarly for $P_2$ assume a mass of $2\space units$ at point $\textbf{B}$ and mass of $1\space unit$ at point $\textbf{A}$.