To solve for the [tex]\( x \)[/tex]-coordinate of [tex]\( Q \)[/tex], let's use the concept of section formula which is used to find the coordinates of a point that divides a line segment internally in a given ratio.
Given:
- The point [tex]\( R \)[/tex] divides the line segment [tex]\( \overline{PQ} \)[/tex] in the ratio 1:3.
- The [tex]\( x \)[/tex]-coordinate of [tex]\( R \)[/tex] is [tex]\(-1\)[/tex].
- The [tex]\( x \)[/tex]-coordinate of [tex]\( P \)[/tex] is [tex]\(-3\)[/tex].
We need to find the [tex]\( x \)[/tex]-coordinate of [tex]\( Q \)[/tex], denoted as [tex]\( x_Q \)[/tex].
By the section formula, if a point [tex]\( R \)[/tex] divides the line segment [tex]\( \overline{PQ} \)[/tex] in the ratio [tex]\( m:n \)[/tex], then the coordinates of [tex]\( R \)[/tex] can be found using:
[tex]\[ \text{Coordinate of } R = \frac{nx_1 + mx_2}{m + n} \][/tex]
Here, [tex]\( x_1 = -3 \)[/tex] (the [tex]\( x \)[/tex]-coordinate of [tex]\( P \)[/tex]), [tex]\( x_2 = x_Q \)[/tex] (the [tex]\( x \)[/tex]-coordinate of [tex]\( Q \)[/tex]), and the point [tex]\( R \)[/tex] is dividing the segment [tex]\( \overline{PQ} \)[/tex] in the ratio [tex]\( 1:3 \)[/tex]. Thus, [tex]\( m = 1 \)[/tex] and [tex]\( n = 3 \)[/tex].
The [tex]\( x \)[/tex]-coordinate of [tex]\( R \)[/tex] is already given as [tex]\( -1 \)[/tex]. Plugging these values into the formula, we get:
[tex]\[ -1 = \frac{3(-3) + 1(x_Q)}{1 + 3} \][/tex]
Now, solve for [tex]\( x_Q \)[/tex]:
[tex]\[ -1 = \frac{3(-3) + x_Q}{4} \][/tex]
First, multiply both sides by 4 to isolate the numerator:
[tex]\[ -1 \times 4 = 3(-3) + x_Q \][/tex]
[tex]\[ -4 = -9 + x_Q \][/tex]
Next, add 9 to both sides to solve for [tex]\( x_Q \)[/tex]:
[tex]\[ -4 + 9 = x_Q \][/tex]
[tex]\[ 5 = x_Q \][/tex]
Therefore, the [tex]\( x \)[/tex]-coordinate of [tex]\( Q \)[/tex] is [tex]\( 5 \)[/tex].
The correct answer is [tex]\( \boxed{5} \)[/tex].
