The equations of two sides AB and AC of a triangle ABC are 4x + y = 14 and 3x – 2y = 5, respectively. The point divides the third side BC internally in the ratio 2:1. the equation of the side BC is

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Mathematics

Introduction to Determinants

Question

The equations of two sides AB and AC of a triangle ABC are 4x + y = 14 and 3x – 2y = 5, respectively. The point $(2, - \frac{4}{3})$ divides the third side BC internally in the ratio 2:1. the equation of the side BC is

x – 3y – 6 = 0

x + 3y + 2 = 0

х – бу – 10 = 0

x + 6y + 6 = 0

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B = (x₂, 14 – 4x₂)

$C = (x_{3}, \frac{3 x_{3} - 5}{2})$

Let $D = (2, \frac{- 4}{3})$

Solving $\frac{2 x_{3} + x_{2}}{3} = 2 \to (1) \frac{3 x_{3} - 5 + 14 - 4 x_{2}}{3} = \frac{- 4}{3} \to (2)$

Slope of BC $= \frac{- 1 + \frac{4}{3}}{1 - 2} = \frac{- 1}{3}$

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