Question

A(7, −1), B(4, 1) and C(−3, 4) are the vertices of a triangle ABC. Find the equation of a line through the vertex B and the point P in AC; such that AP : CP = 2 : 3.

Answer 1

P divides AC in the ratio of 2 : 3

∴ Co-ordinates of P will be

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

`((2(-3) + 3(7))/(2 + 3),(2 xx 4 + 3 xx (-1))/(2 + 3))`

= `((-6 + 21)/5, (8 - 3)/5)`

= `(15/5, 5/5)`

= (3, 1)

∴ Slope of line passing through B and P

= `(y_2 - y_1)/(x_2 - x_1)`

= `(1 - 1)/(3 - 4)`

= `0/(-1)`

= 0

∴ Equation of the required line is given by y – y₁ = m(x – x₁)

`=>` y – 1 = 0(x – 4)

`=>` y – 1 = 0

`=>` y = 1