The diagram is given below:

Coordinates of B

2 = (1 + x)/2 [by section formula]

4 = 1 + x

X = 3

–1 = (–4 + y)/2

–2 = (–4 + y)

Y = 2

∴ the coordinates of B(3,2)

Coordinates of C [by section formula]

0 = (1 + x)/2

0 = (1 + x)

x = –1

–1 = (–4 + y)/2

–2 = (–4 + y)

Y = 2

∴ the coordinates of point C are (–1,2)

Now, Area of triangle ABC

= 1/2(x₁(y₂−y₃) + x₂(y₃−y₁) + x₃(y₁−y₂))

= 1/2(1(2–2) + 3(2 + 4)–1(–4–2))

= 1/2(0 + 18 + 6)

= 1/2(24)

= 12 sq unit