On a number line, the directed line segment from [tex]$Q$[/tex] to [tex]$S$[/tex] has endpoints [tex]$Q$[/tex] at -14 and [tex]$S$[/tex] at 2. Point [tex]$R$[/tex] partitions the directed line segment from [tex]$Q$[/tex] to [tex]$S$[/tex] in a 3:5 ratio.

Which expression correctly uses the formula [tex]$\left(\frac{m}{m+n}\right)\left(x_2-x_1\right)+x_1$[/tex] to find the location of point [tex]$R$[/tex]?

A. [tex]$\left(\frac{3}{3+5}\right)(2-(-14))+(-14)$[/tex]
B. [tex]$\left(\frac{3}{3+5}\right)(-14-2)+2$[/tex]
C. [tex]$\left(\frac{3}{3+5}\right)(2-14)+14$[/tex]
D. [tex]$\left(\frac{3}{3+5}\right)(-14-2)-2$[/tex]



Answer :

To find the location of point [tex]\( R \)[/tex] that partitions the directed line segment from [tex]\( Q \)[/tex] to [tex]\( S \)[/tex] in a [tex]\( 3:5 \)[/tex] ratio, we can use the formula:

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

Given:
- [tex]\( x_1 = -14 \)[/tex] (the coordinate of point [tex]\( Q \)[/tex])
- [tex]\( x_2 = 2 \)[/tex] (the coordinate of point [tex]\( S \)[/tex])
- The ratio [tex]\( 3:5 \)[/tex] translates to [tex]\( m = 3 \)[/tex] and [tex]\( n = 5 \)[/tex]

Substitute these values into the formula:

[tex]\[ R = \left(\frac{3}{3 + 5}\right)(2 - (-14)) + (-14) \][/tex]

First, calculate the fraction:

[tex]\[ \frac{3}{3 + 5} = \frac{3}{8} \][/tex]

Next, compute the difference in the coordinates:

[tex]\[ 2 - (-14) = 2 + 14 = 16 \][/tex]

Now multiply the fraction by this difference:

[tex]\[ \left(\frac{3}{8}\right) \times 16 = \frac{3 \times 16}{8} = 6 \][/tex]

Finally, add this result to [tex]\( x_1 \)[/tex]:

[tex]\[ 6 + (-14) = 6 - 14 = -8 \][/tex]

Therefore, the location of point [tex]\( R \)[/tex] is:

[tex]\[ -8 \][/tex]

The correct expression that uses the formula to find the location of point [tex]\( R \)[/tex] is:

[tex]\[ \left( \frac{3}{3+5} \right)(2 - (-14)) + (-14) \][/tex]

Which matches the first option:

[tex]\[ \left(\frac{3}{3+5}\right)(2-(-14))+(-14) \][/tex]