Question

JK in the coordinate plane has endpoints with coordinates (-4, 11) and (8,
1)
Suppose J, P, and K are collinear on JK and JP:JK = }. What are the coordinates of P?
The coordinates of P are

Answer 1

Answer:

(0, 7)

Step-by-step explanation:

Given:

J(-4, 11)

K(8, -1)

JP:JK = 1/3

Required:

Coordinates of P

SOLUTION:

Use the formula, $(x, y) = (x_1 + k(x_2 - x_1), y_1 + k(y_2 - y_1))$ to find the coordinates of point P, that partition the segment JK into the ratio 1/3.

Let,

$J(-4, 11) = (x_1, y_1)$

$K(8, -1) = (x_2, y_2)$

$k = \frac{1}{3}$

Thus, plug in the values as follows:

$P(x, y) = (-4 + \frac{1}{3}(8 -(-4)), 11 + \frac{1}{3}(-1 - 11)$

$P(x, y) = (-4 + \frac{1}{3}(12), 11 + \frac{1}{3}(-12)$

$P(x, y) = (-4 + \frac{12}{3}, 11 + \frac{-12}{3})$

$P(x, y) = (-4 + 4, 11 + (-4)$

$P(x, y) = (0, 7)$

The coordinates of point P, are (0, 7)