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.

By section formula,

Co-ordinates of P = $\Big(\dfrac{m$1x2 + m2x1}{m1 + m2}, \dfrac{m1y2 + m2y1}{m1 + m2}\Big)

Substituting values we get,

$P = \Big(\dfrac{2 \times -3 + 3 \times 7}{2 + 3}, \dfrac{2 \times 4 + 3 \times -1}{2 + 3}\Big) \\[1em] = \Big(\dfrac{-6 + 21}{5}, \dfrac{8 - 3}{5}\Big) \\[1em] = \Big(\dfrac{15}{5}, \dfrac{5}{5}\Big) \\[1em] = (3, 1).$

Slope of BP = $\dfrac{1 - 1}{3 - 4} = 0.$

By point-slope form,

Equation of BP,

⇒ y - y₁ = m(x - x₁)

⇒ y - 1 = 0(x - 3)

⇒ y - 1 = 0

⇒ y = 1.

Hence, equation of BP is y = 1.