Point [tex]$R$[/tex] divides [tex]$\overline{PQ}$[/tex] in the ratio [tex]$1:3$[/tex]. If the [tex]$x$[/tex]-coordinate of [tex]$R$[/tex] is [tex]$-1$[/tex] and the [tex]$x$[/tex]-coordinate of [tex]$P$[/tex] is [tex]$-3$[/tex], what is the [tex]$x$[/tex]-coordinate of [tex]$Q$[/tex]?

A. [tex]$-\frac{1}{3}$[/tex]
B. [tex]$3$[/tex]
C. [tex]$5$[/tex]
D. [tex]$6$[/tex]
E. [tex]$-9$[/tex]



Answer :

To find the [tex]\(x\)[/tex]-coordinate of point [tex]\(Q\)[/tex], given that point [tex]\(R\)[/tex] divides the line segment [tex]\(\overline{PQ}\)[/tex] in the ratio [tex]\(1:3\)[/tex] and knowing the [tex]\(x\)[/tex]-coordinates of [tex]\(R\)[/tex] and [tex]\(P\)[/tex], we can use the section formula for internal division.

The section formula for internal division states that if a point [tex]\(R\)[/tex] divides the line segment [tex]\(\overline{PQ}\)[/tex] in the ratio [tex]\(m:n\)[/tex], the [tex]\(x\)[/tex]-coordinate of [tex]\(R\)[/tex] can be given by:

[tex]\[ x_R = \frac{m \cdot x_Q + n \cdot x_P}{m + n} \][/tex]

Here, the given values are:
- [tex]\(x_R = -1\)[/tex]
- [tex]\(x_P = -3\)[/tex]
- The ratio [tex]\(m:n = 1:3\)[/tex], which implies [tex]\(m = 1\)[/tex] and [tex]\(n = 3\)[/tex]

Substituting the known values into the section formula, we have:

[tex]\[ -1 = \frac{1 \cdot x_Q + 3 \cdot (-3)}{1 + 3} \][/tex]

Simplify the denominator:

[tex]\[ -1 = \frac{x_Q + 3(-3)}{4} \][/tex]

Next, simplify inside the numerator:

[tex]\[ -1 = \frac{x_Q - 9}{4} \][/tex]

To eliminate the fraction, multiply both sides by 4:

[tex]\[ 4 \cdot (-1) = x_Q - 9 \][/tex]

[tex]\[ -4 = x_Q - 9 \][/tex]

Solve for [tex]\(x_Q\)[/tex]:

[tex]\[ -4 + 9 = x_Q \][/tex]

[tex]\[ x_Q = 5 \][/tex]

Thus, the [tex]\(x\)[/tex]-coordinate of [tex]\(Q\)[/tex] is [tex]\(5\)[/tex]. Therefore, the correct answer is:

[tex]\[ \boxed{5} \][/tex]