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Directed Line Segments and Modeling

What is the \( y \)-coordinate of the point that divides the directed line segment from \( J \) to \( K \) into a ratio of \( 5:1 \)?

[tex]\[
y = \left(\frac{m}{m+n}\right)(y_2 - y_1) + y_1
\][/tex]

Options:
A. \(-8\)
B. \(-5\)
C. \(0\)
D. [tex]\(4\)[/tex]


Sagot :

To find the \( y \)-coordinate of the point that divides the directed line segment from \( J \) to \( K \) into a ratio of \( 5:1 \), we can use the section formula. The general formula to find a point dividing a segment in the ratio \( m:n \) is:

[tex]\[ v = \left( \frac{m}{m+n} \right)(v_2 - v_1) + v_1 \][/tex]

Here, we know the following from the problem statement:

- \( v_1 = -8 \)
- \( v_2 = 0 \)
- \( m = 5 \)
- \( n = 1 \)

We substitute these values into the formula:

[tex]\[ y = \left( \frac{5}{5+1} \right)(0 - (-8)) + (-8) \][/tex]

Now, let's break this into steps:

1. Calculate the sum of \( m \) and \( n \):

[tex]\[ m + n = 5 + 1 = 6 \][/tex]

2. Calculate the fraction:

[tex]\[ \frac{m}{m+n} = \frac{5}{6} \][/tex]

3. Calculate the difference \( v_2 - v_1 \):

[tex]\[ v_2 - v_1 = 0 - (-8) = 0 + 8 = 8 \][/tex]

4. Multiply the fraction by the difference:

[tex]\[ \frac{5}{6} \times 8 = \frac{5 \times 8}{6} = \frac{40}{6} = \frac{20}{3} \][/tex]

5. Add this result to \( v_1 \):

[tex]\[ \frac{20}{3} + (-8) = \frac{20}{3} - 8 = \frac{20}{3} - \frac{24}{3} = \frac{20 - 24}{3} = -\frac{4}{3} \][/tex]

Thus, the \( y \)-coordinate of the point that divides the directed line segment from \( J \) to \( K \) into a ratio of \( 5:1 \) is:

[tex]\[ -\frac{4}{3} \approx -1.33 \][/tex]

Correct to more decimal places, the value is [tex]\( -1.333333333333333 \)[/tex].
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