Given:

The vertices of ∆ABC are A (1, 4), B (− 3, 2) and C (− 5, − 3).

Now let us find the slopes of ∆ABC.

Slope of AB = \( \dfrac{(2 – 4) }{ (-3-1)}\)

= \( \dfrac{1}{2}\)

Slope of BC = \(\dfrac{(-3 – 2) }{ (-5+3)}\)

= \( \dfrac{5}{2}\)

Slope of CA =\( \dfrac{(4 + 3) }{ (1 + 5)}\)

= \( \dfrac{7}{6}\)

Thus, we have:

Slope of CF = -2

Slope of AD = \( \dfrac{-2}{5}\)

Slope of BE = \( \dfrac{-6}{7}\)

Hence,

Equation of CF is:

y + 3 = -2(x + 5)

y + 3 = -2x – 10

2x + y + 13 = 0

Equation of AD is:

y – 4 = (\( \dfrac{-2}{5}\)) (x – 1)

5y – 20 = -2x + 2

2x + 5y – 22 = 0

Equation of BE is:

y – 2 = (\( \dfrac{-6}{7}\)) (x + 3)

7y – 14 = -6x – 18

6x + 7y + 4 = 0

The required equations are 2x + y + 13 = 0, 2x + 5y – 22 = 0, 6x + 7y + 4 = 0.

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