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On a number line, the directed line segment from [tex]\( Q \)[/tex] to [tex]\( S \)[/tex] has endpoints [tex]\( Q \)[/tex] at -2 and [tex]\( S \)[/tex] at 6. Point [tex]\( R \)[/tex] partitions the directed line segment from [tex]\( Q \)[/tex] to [tex]\( S \)[/tex] in a [tex]\( 3:2 \)[/tex] ratio. Rachel uses the section formula to find the location of point [tex]\( R \)[/tex] on the number line. Her work is shown below.

Let [tex]\( m = 3, n = 2, x_1 = -2 \)[/tex], and [tex]\( x_2 = 6 \)[/tex].

1. [tex]\( R = \frac{mx_2 + nx_1}{m+n} \)[/tex]
2. [tex]\( R = \frac{3(6) + 2(-2)}{3+2} \)[/tex]

What is the location of point [tex]\( R \)[/tex] on the number line?

A. [tex]\( \frac{14}{5} \)[/tex]
B. [tex]\( \frac{16}{5} \)[/tex]
C. [tex]\( \frac{18}{5} \)[/tex]
D. [tex]\( \frac{22}{5} \)[/tex]

Sagot :

GinnyAnswer
For the problem at hand, we need to find the location of point [tex]\( R \)[/tex] on the number line. Point [tex]\( R \)[/tex] partitions the directed line segment from [tex]\( Q \)[/tex] to [tex]\( S \)[/tex] in a [tex]\( 3:2 \)[/tex] ratio.

The section formula for a point dividing a line segment internally in a given ratio is given by:
[tex]\[ R = \frac{m x_2 + n x_1}{m + n} \][/tex]

Given:
- [tex]\( m = 3 \)[/tex]
- [tex]\( n = 2 \)[/tex]
- [tex]\( x_1 = -2 \)[/tex] (coordinate of point [tex]\( Q \)[/tex])
- [tex]\( x_2 = 6 \)[/tex] (coordinate of point [tex]\( S \)[/tex])

Let's substitute these values into the section formula:
[tex]\[ R = \frac{3 \cdot 6 + 2 \cdot (-2)}{3 + 2} \][/tex]

Now, simplify this step-by-step:
1. Compute the products in the numerator:
[tex]\[ 3 \cdot 6 = 18 \][/tex]
[tex]\[ 2 \cdot (-2) = -4 \][/tex]

2. Substitute these results back into the formula:
[tex]\[ R = \frac{18 + (-4)}{5} \][/tex]

3. Simplify the expression inside the numerator:
[tex]\[ 18 + (-4) = 14 \][/tex]

4. Divide by the denominator:
[tex]\[ R = \frac{14}{5} \][/tex]

Thus, the location of point [tex]\( R \)[/tex] on the number line is:
[tex]\[ \frac{14}{5} \][/tex]

So, the correct answer is:
[tex]\[ \boxed{\frac{14}{5}} \][/tex]
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