Question

1. fill in the blanks using the available answer choices. find d on \\(\overline{af}\\) that is \\(\frac{3}{5}\\) the distance from a to f. point d is located at \\(\underline{\quad}\\) (blank1) blank1 options: (0,1), (2,3), (4,4), (4,5)

Explanation:

Step1: Identify coordinates of A and F

Point \( A(-2, -1) \), Point \( F(8, 9) \).

Step2: Use section formula (internal division)

The section formula for a point \( D(x, y) \) dividing the line segment joining \( A(x_1, y_1) \) and \( F(x_2, y_2) \) in the ratio \( m:n \) is \( x=\frac{mx_2 + nx_1}{m + n} \), \( y=\frac{my_2 + ny_1}{m + n} \). Here, \( m = 3 \), \( n = 2 \) (since \( AD:DF = 3:2 \), total ratio \( 3 + 2 = 5 \)? Wait, no: Wait, the problem says \( \frac{3}{5} \) the distance from A to F? Wait, no, "that is \( \frac{3}{5} \) the distance from A to F" – so the ratio \( AD:AF = 3:5 \), so \( AD:DF = 3:2 \). So \( m = 3 \), \( n = 2 \).

So \( x=\frac{3\times8 + 2\times(-2)}{3 + 2}=\frac{24 - 4}{5}=\frac{20}{5}=4 \)

\( y=\frac{3\times9 + 2\times(-1)}{3 + 2}=\frac{27 - 2}{5}=\frac{25}{5}=5 \)? Wait, no, wait the options have (4,5)? Wait, but let's check again. Wait, maybe the ratio is \( AD:DF = 3:2 \), so using section formula:

\( x = x_1 + \frac{3}{5}(x_2 - x_1) \), \( y = y_1 + \frac{3}{5}(y_2 - y_1) \)

\( x = -2 + \frac{3}{5}(8 - (-2)) = -2 + \frac{3}{5}(10) = -2 + 6 = 4 \)

\( y = -1 + \frac{3}{5}(9 - (-1)) = -1 + \frac{3}{5}(10) = -1 + 6 = 5 \). Wait, but the options have (4,5) as one of them. Wait, but let's check the options:

Blank1 options: (0,1), (2,3), (4,4), (4,5)

Wait, my calculation gives (4,5). Let's verify:

Vector from A to F: \( (8 - (-2), 9 - (-1)) = (10, 10) \). \( \frac{3}{5} \) of that vector is \( (6, 6) \). So adding to A: \( (-2 + 6, -1 + 6) = (4, 5) \). Yes, that's correct.

Answer:

(4, 5)