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Tia lives at the corner of 4th Street and 8th Avenue. Lei lives at the corner of 12th Street and 20th Avenue. The fruit market is [tex]\(\frac{3}{4}\)[/tex] the distance from Tia's home to Lei's home.

[tex]\[
\begin{array}{l}
x=\left(\frac{m}{m+n}\right)\left(x_2-x_1\right)+x_1 \\
y=\left(\frac{m}{m+n}\right)\left(y_2-y_1\right)+y_1
\end{array}
\][/tex]

Where is the fruit market?

A. 6th Street and 11th Avenue

B. 10th Street and 17th Avenue

C. 9th Street and 15th Avenue

D. 8th Street and 14th Avenue


Sagot :

Let's analyze the problem step by step.

1. Identifying Coordinates:
- Tia's home coordinates: (4, 8)
- Lei's home coordinates: (12, 20)

2. Fractional Distance:
The fruit market is [tex]\(\frac{3}{4}\)[/tex] the distance from Tia's home to Lei's home. This can be understood as splitting the distance into a ratio of [tex]\( \frac{m}{m+n} = \frac{3}{4} \)[/tex]. Using this ratio:
- [tex]\(m = 3\)[/tex]
- [tex]\(n = 1\)[/tex] (since [tex]\(4 - 3 = 1\)[/tex])

3. Formula Application:
Using the given formulas,
[tex]\[ x = \left(\frac{m}{m+n}\right)\left(x_2 - x_1\right) + x_1 \][/tex]
[tex]\[ y = \left(\frac{m}{m+n}\right)\left(y_2 - y_1\right) + y_1 \][/tex]

4. Substitute the Known Values:
- For [tex]\(x\)[/tex]:
[tex]\[ x = \left(\frac{3}{3+1}\right) \left(12 - 4\right) + 4 \][/tex]
Simplifying inside the parentheses:
[tex]\[ x = \left(\frac{3}{4}\right) \left(8\right) + 4 \][/tex]
[tex]\[ x = 6 + 4 = 10 \][/tex]

- For [tex]\(y\)[/tex]:
[tex]\[ y = \left(\frac{3}{3+1}\right) \left(20 - 8\right) + 8 \][/tex]
Simplifying inside the parentheses:
[tex]\[ y = \left(\frac{3}{4}\right) \left(12\right) + 8 \][/tex]
[tex]\[ y = 9 + 8 = 17 \][/tex]

Therefore, the coordinates of the fruit market are (10, 17).

Among the given options:

- 10th Street and 17th Avenue

Thus, the fruit market is at 10th Street and 17th Avenue.
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