Answer :
To find the coordinates of point [tex]\( P \)[/tex] that partitions line segment [tex]\( AB \)[/tex] into a part-to-whole ratio of [tex]\( 1:5 \)[/tex], we can use the section formula for internal division.
Given the endpoints [tex]\( A(-9, 3) \)[/tex] and [tex]\( B(1, 8) \)[/tex], and the ratio [tex]\( 1:5 \)[/tex], let's identify the coordinates of point [tex]\( P \)[/tex].
The section formula states:
[tex]\[ P = \left(\frac{m \cdot x_B + n \cdot x_A}{m + n}, \frac{m \cdot y_B + n \cdot y_A}{m + n}\right) \][/tex]
where [tex]\( (x_A, y_A) = (-9, 3) \)[/tex], [tex]\( (x_B, y_B) = (1, 8) \)[/tex], and the ratio [tex]\( m:n = 1:5 \)[/tex].
Let's fill in these values step-by-step:
1. Identify the coordinates and the ratio:
[tex]\[ A = (-9, 3) \][/tex]
[tex]\[ B = (1, 8) \][/tex]
[tex]\[ m = 1, \, n = 5 \][/tex]
2. Formula for the x-coordinate of [tex]\( P \)[/tex]:
[tex]\[ P_x = \frac{m \cdot x_B + n \cdot x_A}{m + n} = \frac{1 \cdot 1 + 5 \cdot -9}{1 + 5} \][/tex]
Now we calculate:
[tex]\[ P_x = \frac{1 \cdot 1 + 5 \cdot -9}{1 + 5} = \frac{1 - 45}{6} = \frac{-44}{6} = -\frac{22}{3} \approx -7.333333333333333 \][/tex]
3. Formula for the y-coordinate of [tex]\( P \)[/tex]:
[tex]\[ P_y = \frac{m \cdot y_B + n \cdot y_A}{m + n} = \frac{1 \cdot 8 + 5 \cdot 3}{1 + 5} \][/tex]
Now we calculate:
[tex]\[ P_y = \frac{1 \cdot 8 + 5 \cdot 3}{1 + 5} = \frac{8 + 15}{6} = \frac{23}{6} \approx 3.8333333333333335 \][/tex]
Therefore, the coordinates of point [tex]\( P \)[/tex] are [tex]\( \left( -7.333333333333333, 3.8333333333333335 \right) \)[/tex].
So, filling in the correct values for the formula:
[tex]\[ P = \left( \frac{1 \cdot 1 + 5 \cdot (-9)}{1 + 5}, \frac{1 \cdot 8 + 5 \cdot 3}{1 + 5} \right) = \left( \frac{1 + (-45)}{6}, \frac{8 + 15}{6} \right) = \left( -7.333333333333333, 3.8333333333333335 \right) \][/tex]
Finally, the filled formula should be:
[tex]\[ P = \left (\frac{1 \cdot 1 + 5(-9)}{1 + 5}, \frac{1 \cdot 8 + 5(3)}{1 + 5} \right) \][/tex]
Given the endpoints [tex]\( A(-9, 3) \)[/tex] and [tex]\( B(1, 8) \)[/tex], and the ratio [tex]\( 1:5 \)[/tex], let's identify the coordinates of point [tex]\( P \)[/tex].
The section formula states:
[tex]\[ P = \left(\frac{m \cdot x_B + n \cdot x_A}{m + n}, \frac{m \cdot y_B + n \cdot y_A}{m + n}\right) \][/tex]
where [tex]\( (x_A, y_A) = (-9, 3) \)[/tex], [tex]\( (x_B, y_B) = (1, 8) \)[/tex], and the ratio [tex]\( m:n = 1:5 \)[/tex].
Let's fill in these values step-by-step:
1. Identify the coordinates and the ratio:
[tex]\[ A = (-9, 3) \][/tex]
[tex]\[ B = (1, 8) \][/tex]
[tex]\[ m = 1, \, n = 5 \][/tex]
2. Formula for the x-coordinate of [tex]\( P \)[/tex]:
[tex]\[ P_x = \frac{m \cdot x_B + n \cdot x_A}{m + n} = \frac{1 \cdot 1 + 5 \cdot -9}{1 + 5} \][/tex]
Now we calculate:
[tex]\[ P_x = \frac{1 \cdot 1 + 5 \cdot -9}{1 + 5} = \frac{1 - 45}{6} = \frac{-44}{6} = -\frac{22}{3} \approx -7.333333333333333 \][/tex]
3. Formula for the y-coordinate of [tex]\( P \)[/tex]:
[tex]\[ P_y = \frac{m \cdot y_B + n \cdot y_A}{m + n} = \frac{1 \cdot 8 + 5 \cdot 3}{1 + 5} \][/tex]
Now we calculate:
[tex]\[ P_y = \frac{1 \cdot 8 + 5 \cdot 3}{1 + 5} = \frac{8 + 15}{6} = \frac{23}{6} \approx 3.8333333333333335 \][/tex]
Therefore, the coordinates of point [tex]\( P \)[/tex] are [tex]\( \left( -7.333333333333333, 3.8333333333333335 \right) \)[/tex].
So, filling in the correct values for the formula:
[tex]\[ P = \left( \frac{1 \cdot 1 + 5 \cdot (-9)}{1 + 5}, \frac{1 \cdot 8 + 5 \cdot 3}{1 + 5} \right) = \left( \frac{1 + (-45)}{6}, \frac{8 + 15}{6} \right) = \left( -7.333333333333333, 3.8333333333333335 \right) \][/tex]
Finally, the filled formula should be:
[tex]\[ P = \left (\frac{1 \cdot 1 + 5(-9)}{1 + 5}, \frac{1 \cdot 8 + 5(3)}{1 + 5} \right) \][/tex]
