Question

The line joining the points (2, 1) and (5, –8) is trisected at the point P and Q. If point P lies on the line 2x – y + k = 0, find the value of k. Also, find the co-ordinates of point Q.

Answer 1

Let A(2, 1) and B(5, –8) be the given point trisected by the points P and Q 

`=>` AP = PQ = QB 

For P:

m₁ : m₂ = AP : PB = 1 : 2

(x₁, y₁) = (2, 1) and (x₂, y₂) = (5, 8)

∴ `x = (1 xx 5 + 2 xx 2)/(1 + 2)`

= `(5 + 4)/3`

= `9/3`

= 3 

`y = (1 xx (-8) + 2 xx 1)/(1 + 2)`

= `(-8 + 2)/3`

= `-6/3`

= –2

∴ Coordinates of P are (3, 2)

Since point P line on the line 2x – y + k = 0

2(3) – (–2) + k = 0

`=>` 6 + 2 + k = 0 

For Q:

m₁ : m₂ = AQ : QB = 2 : 1

(x₁, y₁) = (2, 1) and (x₂, y₂) = (5, –8) 

∴ `x = (2 xx 5 + 1 xx 2)/(2 + 1)`

= `(10 + 2)/3`

= `12/3`

= 4

`y = (2 xx (-8) + 1 xx 1)/(1 + 2)`

= `(-16 + 1)/3`

= `(-15)/3`

= –5

∴ Coordinates of Q are (4, –5)