What is the [tex]$x$[/tex]-coordinate of the point that divides the directed line segment from [tex]$K$[/tex] to [tex]$J$[/tex] into a ratio of 1:3?

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

A. \(-1\)
B. \(3\)
C. \(7\)
D. [tex]\(11\)[/tex]



Answer :

To determine the \( x \)-coordinate of the point that divides the directed line segment joining points \( K \) and \( J \) in the ratio \( 1:3 \), we will use the section formula for the \( x \)-coordinate. The formula is given by:

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

Here, \( K \) and \( J \) are points on the coordinate plane with \( K \) having coordinates \( (x_1, y_1) \) and \( J \) having coordinates \( (x_2, y_2) \). The ratio in which the line segment is divided is given by \( m:n \).

Given:
- \( K \) has the coordinates \( (x_1, y_1) = (-1, y_1) \)
- \( J \) has the coordinates \( (x_2, y_2) = (3, y_2) \)
- The ratio \( m:n = 1:3 \)

Let's plug in these values into the formula:

[tex]\[ x_1 = -1 \][/tex]
[tex]\[ x_2 = 3 \][/tex]
[tex]\[ m = 1 \][/tex]
[tex]\[ n = 3 \][/tex]

Now substitute these values into the formula:

[tex]\[ x = \left(\frac{1}{1+3}\right)\left(3 - (-1)\right) + (-1) \][/tex]
[tex]\[ x = \left(\frac{1}{4}\right) \times (3 + 1) + (-1) \][/tex]
[tex]\[ x = \left(\frac{1}{4}\right) \times 4 + (-1) \][/tex]
[tex]\[ x = 1 + (-1) \][/tex]
[tex]\[ x = 0 \][/tex]

Therefore, the [tex]\( x \)[/tex]-coordinate of the point that divides the directed line segment from [tex]\( K \)[/tex] to [tex]\( J \)[/tex] into a ratio of [tex]\( 1:3 \)[/tex] is [tex]\( \boxed{0} \)[/tex].