Question

Consider line segment AB.

A: (-4, 10)
B: (6,2)
Find the coordinates of point P such that AP: PB is 2:3. Enter your answer as a DECIMAL, if
needed.

Answer 1

P(0, 6.8)

Explanation:

• Let (x, y) be the required coordinates of point P.

• 
  `A(-4, \: 10) =(x_1, \: y_1)\: & \: B(6, \: 2) =(x_2,\: y_2)`

• 
  `AP : PB = 2:3 \implies m = 2 \: & \: n = 3`

• Now, by section formula of internal division, we have.

• 
  `x=(mx_2+nx_1)/(m+n) \: & \: y=(my_2+ny_1)/(m+n)`

• 
  `\implies x=(2(6)+3(-4))/(2+3) \: & \: y=(2(2)+3(10))/(2+3)`

• 
  `\implies x=(12-12)/(5) \: & \: y=(4+30)/(5)`

• 
  `\implies x=(0)/(5) \: & \: y=(34)/(5)`

• 
  `\implies x=0 \: & \: y=6.8`

• 
  `\implies P(x,\:y) = P(0,\: 6.8)`

• Thus, the coordinates of point P are (0, 6.8).