Question

Find the area of triangle ABC with A(1, –4) and the mid-points of sides through A being (2, –1) and (0, –1).

Answer 1

Let the coordinates of the points B and C be (x, y) and (a, b), then `(x + 1)/2 = 2`

⇒ x = 4 – 1 = 3 and `(y - 4)/2 = -1`

⇒ y = –2 + 4

⇒ y = 2

Similarly, `(a + 1)/2 = 0`

⇒ a = –1

And `(b - 4)/2 = -1`

⇒ b = –2 + 4 = 2

So, the coordinates of B and Care (3, 2) and (–1, 2)

Here, x₁ = 1, y₁ = –4, x₂ = 3, y₂ = 2 and x₃ = –1, y₃ = 2

∴ Area of ΔABC = `1/2 [x_1 (y_2 - y_3) + x_2 (y_3 - y_1) + x_3 (y_1 - y_2)]`

= `1/2 [1(2 - 2) + 3(2 + 4) - 1(-4 - 2)]`

= `1/2 [0 + 18 + 6]`

= `1/2 xx 24`

= 12 square units.