Question

The points A(x₁, y₁)_, B(x₂, y₂) and C(x₃, y₃) are the vertices of ∆ABC. Find the coordinates of the point P on AD such that AP : PD = 2 : 1

Answer 1

According to the question,

The vertices of ΔABC = A, B and C

Coordinates of A, B and C = A(x₁, y₁), B(x₂, y₂), C(x₃, y₃)

Let the coordinates of a point P be (x, y)

Given,

The ratio in which the point P(x, y), divide the line joining,

`"A"(x_1, y_1)` and `"D"((x_2 + x_3)/2, (y_2 + y_3)/2)` = 2 : 1

Then,

Coordinates of P = `[(2 xx  ((x_2 + x_3)/2) + 1 xx x_1)/(2 + 1), (2 xx ((y_2 + y_3)/2) + 1 xx  y_1)/(2 + 1)]`

By using internal section formula;

= `((m_1x_2 + m_2x_1)/(m_1 + m_2), (m_1y_2 + m_2y_1)/(m_1 + m_2))`

= `((x_2 + x_3 + x_1)/3, (y_2 + y_3 + y_1)/3)`