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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 [tex]$1:2$[/tex]?

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

A. [tex]$(0,1)$[/tex]
B. [tex]$(-1,3)$[/tex]
C. [tex]$(-2,5)$[/tex]
D. [tex]$(1,0)$[/tex]

Sagot :

GinnyAnswer
To find the coordinates of point [tex]\(E\)[/tex] that partitions the directed line segment from [tex]\(A\)[/tex] to [tex]\(B\)[/tex] into a ratio of [tex]\(1:2\)[/tex], let's follow these steps systematically.

Given the coordinates of points [tex]\(A\)[/tex] and [tex]\(B\)[/tex] as:
- [tex]\(A (0, 1)\)[/tex]
- [tex]\(B (-1, 3)\)[/tex]

We are given the ratio [tex]\(1:2\)[/tex], where [tex]\(m = 1\)[/tex] and [tex]\(n = 2\)[/tex].

We will use the section formula for internal division to find the coordinates of point [tex]\(E\)[/tex] that partitions the line segment [tex]\(AB\)[/tex] in the given ratio.

The section formula is:

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

Substitute the known values into the formulas:

Coordinates of [tex]\(A\)[/tex] are [tex]\( (x_1, y_1) = (0, 1) \)[/tex]
Coordinates of [tex]\(B\)[/tex] are [tex]\( (x_2, y_2) = (-1, 3) \)[/tex]
Ratio [tex]\(m:n = 1:2\)[/tex] so [tex]\(m = 1, n = 2\)[/tex]

First, calculate the [tex]\(x\)[/tex]-coordinate of point [tex]\(E\)[/tex]:

[tex]\[ x = \left(\frac{1}{1+2}\right) \left(x_2 - x_1\right) + x_1 \][/tex]
[tex]\[ x = \left(\frac{1}{3}\right) \left( -1 - 0 \right) + 0 \][/tex]
[tex]\[ x = \left(\frac{1}{3}\right) (-1) \][/tex]
[tex]\[ x = -\frac{1}{3} \][/tex]

Next, calculate the [tex]\(y\)[/tex]-coordinate of point [tex]\(E\)[/tex]:

[tex]\[ y = \left(\frac{1}{1+2}\right) \left(y_2 - y_1\right) + y_1 \][/tex]
[tex]\[ y = \left(\frac{1}{3}\right) (3 - 1) + 1 \][/tex]
[tex]\[ y = \left(\frac{1}{3}\right) (2) + 1 \][/tex]
[tex]\[ y = \frac{2}{3} + 1 \][/tex]
[tex]\[ y = \frac{2}{3} + \frac{3}{3} \][/tex]
[tex]\[ y = \frac{5}{3} \][/tex]

Therefore, the coordinates of point [tex]\(E\)[/tex] that partitions the directed line segment from [tex]\(A\)[/tex] to [tex]\(B\)[/tex] into a ratio of [tex]\(1:2\)[/tex] are:

[tex]\[ \left( -\frac{1}{3}, \frac{5}{3} \right) \][/tex]

Approximating these values,

[tex]\[ \left( -0.3333333333333333, 1.6666666666666665 \right) \][/tex]

Therefore, the [tex]\(x\)[/tex]- and [tex]\(y\)[/tex]-coordinates of point [tex]\(E\)[/tex] are [tex]\(-0.3333\)[/tex] and [tex]\(1.6667\)[/tex] respectively.
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