What are the [tex]$x$[/tex]- and [tex]$y$[/tex]-coordinates of point [tex]$E$[/tex], which partitions the directed line segment from [tex]$J$[/tex] to [tex]$K$[/tex] into a ratio of [tex]$1:4$[/tex]?

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

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

A. [tex]$(-13, -3)$[/tex]
B. [tex]$(-7, -1)$[/tex]
C. [tex]$(-5, 0)$[/tex]
D. [tex]$(17, 11)$[/tex]



Answer :

To find the coordinates of point [tex]\( E \)[/tex] which partitions the directed line segment from [tex]\( J \)[/tex] to [tex]\( K \)[/tex] into a ratio of [tex]\( 1: 4 \)[/tex], we will use the section formula. The section formula helps determine the coordinates of a point dividing a line segment in a given ratio.

Given:
- Coordinates of point [tex]\( J \)[/tex] are [tex]\((x_1, y_1) = (-13, -3)\)[/tex].
- Coordinates of point [tex]\( K \)[/tex] are [tex]\((x_2, y_2) = (-7, -1)\)[/tex].
- The ratio [tex]\( m:n = 1:4 \)[/tex].

We will utilize the section formulas for the [tex]\( x \)[/tex]- and [tex]\( y \)[/tex]-coordinates:
[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]

Let's solve for the [tex]\( x \)[/tex]-coordinate first:
[tex]\[ x = \left(\frac{1}{1+4}\right)(-7 - (-13)) + (-13) \][/tex]
[tex]\[ x = \left(\frac{1}{5}\right)(-7 + 13) - 13 \][/tex]
[tex]\[ x = \left(\frac{1}{5}\right)(6) - 13 \][/tex]
[tex]\[ x = 1.2 - 13 \][/tex]
[tex]\[ x = -11.8 \][/tex]

Next, let's solve for the [tex]\( y \)[/tex]-coordinate:
[tex]\[ y = \left(\frac{1}{1+4}\right)(-1 - (-3)) + (-3) \][/tex]
[tex]\[ y = \left(\frac{1}{5}\right)(-1 + 3) - 3 \][/tex]
[tex]\[ y = \left(\frac{1}{5}\right)(2) - 3 \][/tex]
[tex]\[ y = 0.4 - 3 \][/tex]
[tex]\[ y = -2.6 \][/tex]

Therefore, the coordinates of point [tex]\( E \)[/tex] are:
[tex]\[ (-11.8, -2.6) \][/tex]