Division of a segment in a given ratio (vector and coordinate methods).

Knowing the coordinates of points $M_1(x_1,y_1,z_1)$ and $M_2(x_2,y_2,z_2)$ and the ratio $\lambda$ at which the point $M$ divides the directed segment $\overline{M_1{M}_2}$, we will find the coordinates of point $M$.

Let $O$ be the origin. Let's denote $\overline{O{M}_1}=r_1$, $\overline{O{M}_2}=r_2$, $\overline{OM}=r$. Since $$\overline{M_{1}M}=r-r_1,\overline{M{M}_2}=r_2-r,$$ then $r-r_1=\lambda (r_2-r)$, from which (since $\lambda \ne -1$) $$r=\frac{r_1+\lambda r_2}{1+\lambda }.$$

This obtained form provides the solution to the problem in vector form. Transitioning to coordinates in this formula, we get$$x=\frac{x_1+\lambda x_2}{1+\lambda },y=\frac{y_1+\lambda y_2}{1+\lambda },z=\frac{z_1+\lambda z_2}{1+\lambda }.$$

Examples.

2.57. The segment with endpoints at $A(3,-2)$ and $B(6,4)$ is divided into three equal parts. Find the coordinates of the division points.

Solution.

Let $C(x_C,y_C)$ and $D(x_D,y_D)$ be the points dividing the segment $AB$ into three equal parts. Then $${\lambda }_1=\frac{AC}{CB}=\frac{1}{2};$$ $$x_C=\frac{x_A+{\lambda }_1{x}_B}{1+{\lambda }_1}=\frac{3+\frac{1}{2}\cdot 6}{1+\frac{1}{2}}=4;$$

$$y_C=\frac{y_A+{\lambda }_1{y}_B}{1+{\lambda }_1}=\frac{-2+\frac{1}{2}\cdot 4}{1+\frac{1}{2}}=0.$$

Next, we find the coordinates of point $D$:

$${\lambda }_2=\frac{AD}{DB}=\frac{2}{1}=2;$$ $$x_D=\frac{x_A+{\lambda }_2{x}_B}{1+{\lambda }_2}=\frac{3+2\cdot 6}{1+2}=5;$$

$$y_D=\frac{y_A+{\lambda }_2{y}_B}{1+{\lambda }_2}=\frac{-2+2\cdot 4}{1+2}=2.$$

Answer: $(4,0)$ и $(5,2).$

2.58. Determine the coordinates of the endpoints of a segment that is divided into three equal parts, given that the points are $C(2,0,2)$ and $D(5,-2,0)$.

Solution:

Let $A(x_A,y_A,z_A)$ and $B(x_B,y_B,z_B)$ be the endpoints of the given segment.

We'll write down the formulas for finding the coordinates of point $C$ and substitute the known coordinates:

$${\lambda }_1=\frac{AC}{CB}=\frac{1}{2};$$ $$x_C=\frac{x_A+{\lambda }_1{x}_B}{1+{\lambda }_1}\Rightarrow 2=\frac{x_A+\frac{1}{2}\cdot x_B}{1+\frac{1}{2}}=2\frac{x_A+\frac{1}{2}\cdot x_B}{3}\Rightarrow$$ $$\Rightarrow 3=x_A+\frac{1}{2}\cdot x_B;$$

$$y_C=\frac{y_A+{\lambda }_1{y}_B}{1+{\lambda }_1}\Rightarrow 0=\frac{y_A+\frac{1}{2}\cdot y_B}{1+\frac{1}{2}}\Rightarrow 0=y_A+\frac{1}{2}\cdot y_B;$$

$$z_C=\frac{z_A+{\lambda }_1{z}_B}{1+{\lambda }_1}\Rightarrow 2=\frac{z_A+\frac{1}{2}\cdot z_B}{1+\frac{1}{2}}=2\frac{z_A+\frac{1}{2}\cdot z_B}{3}\Rightarrow$$ $$\Rightarrow 3=z_A+\frac{1}{2}\cdot z_B.$$

Similar equations will be written for point $D$:

$${\lambda }_2=\frac{AD}{DB}=\frac{2}{1}=2;$$ $$x_D=\frac{x_A+{\lambda }_2{x}_B}{1+{\lambda }_2}\Rightarrow 5=\frac{x_A+2\cdot x_B}{1+2}=\frac{x_A+2\cdot x_B}{3}\Rightarrow$$ $$\Rightarrow 15=x_A+2\cdot x_B;$$

$$y_D=\frac{y_A+{\lambda }_2{y}_B}{1+{\lambda }_2}\Rightarrow -2=\frac{y_A+2\cdot y_B}{1+2}\Rightarrow -6=y_A+2\cdot y_B;$$

$$z_D=\frac{z_A+{\lambda }_2{z}_B}{1+{\lambda }_2}\Rightarrow 0=\frac{z_A+2\cdot z_B}{1+2}\Rightarrow 0=z_A+2\cdot z_B.$$

Next, we will write the obtained equations for $x_A,x_B;$ $y_A,y_B;$ and $z_A,z_B$ pairwise as systems and solve them:

$$\{\begin{array}{l}{x}_A+\frac{1}{2}{x}_B=3\\ x_A+2x_B=15\end{array}\Rightarrow \{\begin{array}{l}{x}_A=3-0,5x_B\\ 3-0,5x_B+2x_B=15\end{array}\Rightarrow$$ $$\Rightarrow \{\begin{array}{l}{x}_A=3-0,5\cdot 8=-1\\ x_B=\frac{12}{1,5}=8\end{array}$$

$$\{\begin{array}{l}{y}_A+\frac{1}{2}{y}_B=0\\ y_A+2y_B=-6\end{array}\Rightarrow \{\begin{array}{l}{y}_B=-2y_A\\ y_A-4y_A=-6\end{array}\Rightarrow$$ $$\Rightarrow \{\begin{array}{l}{y}_B=-4\\ y_A=2\end{array}$$

$$\{\begin{array}{l}{z}_A+\frac{1}{2}{z}_B=3\\ z_A+2z_B=0\end{array}\Rightarrow \{\begin{array}{l}-2z_B+0,5z_B=3\\ z_A=-2z_B\end{array}\Rightarrow$$ $$\Rightarrow \{\begin{array}{l}{z}_B=-2\\ z_A=4\end{array}$$

Thus, we obtained the coordinates of the endpoints of the segment $A(-1,2,4)$ and $B(8,-4,-2).$

Answer: $A(-1,2,4),$ $B(8,-4,-2).$

Tags: Division of a segment in a given ratio, algebraic lines, geometry