To solve the problem, we need to use the formula for finding a location that divides a line segment in a given ratio. The formula is:
[tex]\[
\left(\frac{m}{m+n}\right)\left(x_2-x_1\right)+x_1
\][/tex]
Here, \(Q\) is at \(-8\) (which we'll denote as \(x_1\)), and \(S\) is at \(12\) (which we'll denote as \(x_2\)). The ratio provided is \(4:1\), which corresponds to \(m = 4\) and \(n = 1\).
We need to substitute these values into the formula:
[tex]\[
\left(\frac{m}{m+n}\right)\left(x_2-x_1\right)+x_1
\][/tex]
Substituting \(m = 4\), \(n = 1\), \(x_1 = -8\), and \(x_2 = 12\), we get:
[tex]\[
\left(\frac{4}{4+1}\right)(12-(-8))+(-8)
\][/tex]
First, calculate the denominator inside the fraction:
[tex]\[
4 + 1 = 5
\][/tex]
So the formula becomes:
[tex]\[
\left(\frac{4}{5}\right)(12-(-8))+(-8)
\][/tex]
Next, simplify inside the parentheses:
[tex]\[
12 - (-8) = 12 + 8 = 20
\][/tex]
So now the formula is:
[tex]\[
\left(\frac{4}{5}\right)(20)+(-8)
\][/tex]
Calculate the fraction times 20:
[tex]\[
\frac{4}{5} \times 20 = 16
\][/tex]
Add this result to \(-8\):
[tex]\[
16 + (-8) = 8
\][/tex]
Therefore, the location that divides the directed line segment from \(Q\) to \(S\) in a \(4:1\) ratio is:
[tex]\[
8
\][/tex]
The correct expression you need that matches the steps taken is:
[tex]\[
\left(\frac{4}{4+1}\right)(12-(-8))+(-8)
\][/tex]
