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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.

Straight Line Eq

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Answer

By section formula,

Co-ordinates of P = (m1x2+m2x1m1+m2,m1y2+m2y1m1+m2)\Big(\dfrac{m1x2 + m2x1}{m1 + m2}, \dfrac{m1y2 + m2y1}{m1 + m2}\Big)

Substituting values we get,

P=(2×−3+3×72+3,2×4+3×−12+3)=(−6+215,8−35)=(155,55)=(3,1).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 = 1−13−4=0.\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.

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