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What are the [tex]\(x\)[/tex]- and [tex]\(y\)[/tex]-coordinates of point [tex]\(E\)[/tex], which partitions the directed line segment from [tex]\(A\)[/tex] to [tex]\(B\)[/tex] into a ratio of 1:2?

[tex]\[
\begin{array}{l}
x = \left(\frac{m}{m+n}\right)\left(x_2 - x_1\right) + x_1 \\
y = \left(\frac{m}{m+n}\right)\left(y_2 - y_1\right) + y_1
\end{array}
\][/tex]

A. [tex]\((0, 1)\)[/tex]

B. [tex]\((-1, 3)\)[/tex]

C. [tex]\((-2, 5)\)[/tex]

D. [tex]\((1, 0)\)[/tex]



Answer :

GinnyAnswer
To find the coordinates of point [tex]\( E \)[/tex] which divides the directed line segment from [tex]\( A(0,1) \)[/tex] to [tex]\( B(1,0) \)[/tex] in the ratio 1:2, we will use the section formula. The section formula helps to find a point that divides a line segment internally in a given ratio.

Given:
- Coordinates of point [tex]\( A \)[/tex]: [tex]\((0, 1)\)[/tex]
- Coordinates of point [tex]\( B \)[/tex]: [tex]\((1, 0)\)[/tex]
- Ratio: [tex]\(1:2\)[/tex]
- Let [tex]\( m = 1 \)[/tex] and [tex]\( n = 2 \)[/tex]

The section formula for a point [tex]\( E \)[/tex] dividing the segment [tex]\( AB \)[/tex] in the ratio [tex]\( m:n \)[/tex] is:
[tex]\[ E = \left( \frac{mx_2 + nx_1}{m+n}, \frac{my_2 + ny_1}{m+n} \right) \][/tex]

Applying the coordinates:
- [tex]\( x_1 = 0 \)[/tex]
- [tex]\( y_1 = 1 \)[/tex]
- [tex]\( x_2 = 1 \)[/tex]
- [tex]\( y_2 = 0 \)[/tex]

We will calculate the [tex]\( x \)[/tex]-coordinate of point [tex]\( E \)[/tex]:
[tex]\[ x_E = \frac{m \cdot x_2 + n \cdot x_1}{m + n} = \frac{1 \cdot 1 + 2 \cdot 0}{1 + 2} = \frac{1}{3} \][/tex]

Next, we will calculate the [tex]\( y \)[/tex]-coordinate of point [tex]\( E \)[/tex]:
[tex]\[ y_E = \frac{m \cdot y_2 + n \cdot y_1}{m + n} = \frac{1 \cdot 0 + 2 \cdot 1}{1 + 2} = \frac{2}{3} \][/tex]

Therefore, the coordinates of point [tex]\( E \)[/tex] are:
[tex]\[ \left( \frac{1}{3}, \frac{2}{3} \right) \][/tex]

In decimal form, these coordinates are approximately:
[tex]\[ (0.3333333333333333, 0.6666666666666667) \][/tex]

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