Question

The line segment joining the points A(2, 1) and B(5, −8) is trisected at the points P and Q such that P is nearer to A. If P also lies on the line given by 2x − y + k = 0, find the value of k.

Answer 1

Given:

Points: A(2, 1) and B(5, −8)

Line segment AB is trisected at points P and Q

P is nearer to A

P lies on the line: 2x − y + k = 0

Use Section Formula to find coordinates of point P

So point P divides AB in the ratio 1 : 2 (P is closer to A).

`P = ((2x_1 + 1x_2)/(1+2)), (2y_1+1y_2)/(1+2)`

Where:

A = (x₁​, y_1​) = (2, 1)

B = (x₂, y₂) = (5, −8) 

`P_x = (2(2) + 1(5))/3`

`= (4+5)/3`

`=9/3`

= 3

`P_y = (2(1) + 1(-8))/3`

`= (2-8)/3`

`= -6/3`

= −2

P = (3, −2)

2x − y + k = 0

2(3) − (−2) + k = 0

⇒ 6 + 2 + k = 0

⇒ 8 + k = 0

⇒ k = −8