Question

The line segment joining the points (3, −1) and (−6, 5) is trisected. Find the co-ordinates of the points of trisection.

Answer 1

If a point divides the line joining A(x₁, y₁) and B(x₂, y₂) internally in the ratio m₁ : m₂ (where m₁ = AP and m₂ = PB),

Then the coordinates of that point are,

x = `((m_1 xx x_2) + (m_2 xx x_1)) / (m_1 + m_2)`,

y = `((m_1 xx y_2) + (m_2 xx y_1)) / (m_1 + m_2)`

Given:

A = (3, −1) and B = (−6, 5),

Here, let P and Q trisect AB with P nearer A.

⇒ Point P: AP : PB = 1 : 2,

So m₁ = 1, m₂ = 2

xP = `((1 xx x_2) + (2 xx x_1)) / (1 + 2)`

xP = `(1 xx (−6) + 2 xx 3) / 3`

xP = `(−6 + 6) / 3`

∴ xP = 0

yP = `((1 xx y_2) + (2 xx y_1)) / 3`

yP = `(1 xx 5 + 2 xx (−1)) / 3`

yP = `(5 − 2) / 3`

∴ yP = 1

So P = (0, 1)

⇒ Point Q: AQ : QB = 2 : 1,

So m₁ = 2, m₂ = 1

xQ = `((2 xx x_2) + (1 xx x_1)) / (2 + 1)`

xQ = `(2 xx (−6) + 1 xx 3)/3`

xQ = `(−12 + 3) / 3`

∴ xQ = −3

yQ = `((2 xx y_2) + (1 xx y_1))/3`

yQ = `(2 xx 5 + 1 xx (−1)) / 3`

yQ = `(10 − 1)/3`

∴ yQ = 3

So Q = (−3, 3)

Hence, the Points of trisection are P = (0, 1) and Q = (−3, 3).