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

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



Answer :

GinnyAnswer
To find the coordinates of point [tex]\( E \)[/tex] that partitions the directed line segment from point [tex]\( A = (x_1, y_1) \)[/tex] to point [tex]\( B = (x_2, y_2) \)[/tex] in the ratio [tex]\( m:n \)[/tex], we use the section formula. The coordinates of the point [tex]\( E \)[/tex], which divides the line segment joining [tex]\( A \)[/tex] and [tex]\( B \)[/tex] in the ratio [tex]\( m:n \)[/tex], are given by:

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

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

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

Let's plug these values into the formulas:

For the [tex]\( x \)[/tex]-coordinate:

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

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

[tex]\[ x = \frac{1}{3} \][/tex]

For the [tex]\( y \)[/tex]-coordinate:

[tex]\[ y = \left(\frac{1}{1+2}\right)(0 - 1) + 1 \][/tex]

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

[tex]\[ y = -\frac{1}{3} + 1 \][/tex]

[tex]\[ y = \frac{2}{3} \][/tex]

So, the coordinates of point [tex]\( E \)[/tex] that divides the line segment [tex]\( AB \)[/tex] in the ratio 1:2 are:

[tex]\[ \left( \frac{1}{3}, \frac{2}{3} \right) \approx (0.3333333333333333, 0.6666666666666667) \][/tex]

Therefore, the coordinates of point [tex]\( E \)[/tex] are [tex]\( \left( 0.3333333333333333, 0.6666666666666667 \right) \)[/tex].

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