Question

Line segment JK has endpoints with coordinates (-4,11) and (8,-1). J,P, and K are collinear on segment JK, and JP: JK = 1/3. What are the coordinates of P?

Answer 1

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

Explanation:

Given

`J = (-4,11)`

`K = (8,-1)`

`JP:JK = 1/3`

Required

Determine the coordinates of P

`JP:JK = 1/3`

Represent as ratio

`JP:JK = 1:3`

Next, is to determine
`JP:PK`

`PK = JK - JP`

So,

`JP : PK = 1 : 3 -1`

`JP : PK = 1 : 2`

The coordinates of P can be calculated using:

`P(x,y) = ((mx_2 + nx_1)/(m+n),(my_2 + ny_1)/(m+n))`

In this case:

`(x_1,y_1) = (-4,11)`

`(x_2,y_2) = (8,-1)`

`m:n = 1:2`

So:

`P(x,y) = ((mx_2 + nx_1)/(m+n),(my_2 + ny_1)/(m+n))`

`P(x,y) = ((1 * 8 + 2 * -4)/(1 + 2),(1 * -1 + 2 * 11)/(1 + 2))`

`P(x,y) = ((0)/(3),(21)/(3))`

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

Hence, the coordinates of P:
`P(x,y) = (0,7)`