Answer :
Answer:
Step-by-step explanation:
NP:OP is 1:3 1 + 3 = 4
point P is 1/4th of the way from N to O
N is -3,-8 and O is 1, 0
For 'x' from -3 to 1 is a distance of 4 units .... 1/4 of this would be '1'
add this to -3 to get the 'x' coordinate of -2
For 'y' the distance from -8 to 0 is 8 units.... 1/4 of this is 2 units...add this to -8 to get the 'y' coordinate of -6
Point O is then at -2, -6 Plot that on the line for your answer
Answer:
P(-2, -6)
Step-by-step explanation:
Section Formula
Some two points, A(x1, y1) and B(x2, y2), form segment AB.
Let us assume there exists some point C(x3, y3) such that:
[tex]\frac{\text{AC}}{\text{CB}} = \frac kl[/tex]
Then the following is always true about point C's coordinates:
[tex]x_3 = \frac{l \times x_2 + k \times x_1}{l + k}\\\\y_3 = \frac{l \times y_2 + k \times y_1}{l + k}[/tex]
Analyzing The Problem
The two endpoints are N(-3, -8) and O(1, 0).
The point P(x3, y3) lies on the segment NO such that:
[tex]\frac{\text{NP}}{\text{OP}} = \frac13 = \frac kl[/tex]
Therefore, according to the section formula:
[tex]x_3 = \frac{3 \times -3 + 1 \times 1}{3 + 1} = \frac{-9 + 1}4 = \frac{-8}4 = -2\\\\y_3 = \frac{3 \times -8 + 1 \times 0}{3 + 1} = \frac{-24 + 0}4 = \frac{-24}4 = -6[/tex]
The coordinates of point P are (-2, -6).