Question

Find the area of ∆ABC whose vertices are:

(i) A(1, 2), B(−2, 3) and C(−3, −4)
(ii) A(−5, 7), B(−4, −5) and C(4, 5)
(iii) A(3, 8), B(−4, 2) and C(5, −1)
(iv) A(10, −6), B(2, 5) and C(−1, 3)

Answer 1

(i) A(1, 2), B(−2, 3) and C(−3, −4) are the vertices of ∆ABC. Then,
(x₁ = 1, y₁ = 2), (x₂ = −2, y₂ = 3) and (x₃ = −3, y₃ = -4)
$$\begin{gathered} \mathrm{Area}\mathrm{of}\mathrm{triangle}ABC \\ =\frac{1}{2}\left\{x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_3\left(y_1-y_2\right)\right\} \\ =\frac{1}{2}\left\{1\left(3-\left(-4\right)\right)+\left(-2\right)\left(-4-2\right)+\left(-3\right)\left(2-3\right)\right\} \\ =\frac{1}{2}\left\{1\left(3+4\right)-2\left(-6\right)-3\left(-1\right)\right\} \\ =\frac{1}{2}\left\{7+12+3\right\} \\ =\frac{1}{2}\left(22\right) \\ =11\mathrm{sq}.\mathrm{units} \end{gathered}$$

(ii) A(−5, 7), B(−4, −5) and C(4, 5) are the vertices of ∆ABC. Then,
(x₁ = −5, y₁ = 7), (x₂ = −4, y₂ = −5) and (x₃ = 4, y₃ = 5)
$$\begin{gathered} \mathrm{Area}\mathrm{of}\mathrm{triangle}ABC \\ =\frac{1}{2}\left\{x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_3\left(y_1-y_2\right)\right\} \\ =\frac{1}{2}\left\{\left(-5\right)\left(-5-5\right)+\left(-4\right)\left(5-7\right)+4\left(7-\left(-5\right)\right)\right\} \\ =\frac{1}{2}\left\{\left(-5\right)\left(-10\right)-4\left(-2\right)+4\left(12\right)\right\} \\ =\frac{1}{2}\left\{50+8+48\right\} \\ =\frac{1}{2}\left(106\right) \\ =53\mathrm{sq}.\mathrm{units} \end{gathered}$$

(iii) A(3, 8), B(−4, 2) and C(5, −1) are the vertices of ∆ABC. Then,
(x₁ = 3, y₁ = 8), (x₂ = −4, y₂ = 2) and (x₃ = 5, y₃ = −1)
$$\begin{gathered} \mathrm{Area}\mathrm{of}\mathrm{triangle}ABC \\ =\frac{1}{2}\left\{x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_3\left(y_1-y_2\right)\right\} \\ =\frac{1}{2}\left\{3\left(2-\left(-1\right)\right)+\left(-4\right)\left(-1-8\right)+5\left(8-2\right)\right\} \\ =\frac{1}{2}\left\{3\left(2+1\right)-4\left(-9\right)+5\left(6\right)\right\} \\ =\frac{1}{2}\left\{9+36+30\right\} \\ =\frac{1}{2}\left(75\right) \\ =37.5\mathrm{sq}.\mathrm{units} \end{gathered}$$

(iv) A(10, −6), B(2, 5) and C(−1, −3) are the vertices of ∆ABC. Then,
(x₁ = 10, y₁ = −6), (x₂ = 2, y₂ = 5) and (x₃ = −1, y₃ = 3)
$$\begin{gathered} \mathrm{Area}\mathrm{of}\mathrm{triangle}ABC \\ =\frac{1}{2}\left\{x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_3\left(y_1-y_2\right)\right\} \\ =\frac{1}{2}\left\{10\left(5-3\right)+2\left(3-\left(-6\right)\right)+\left(-1\right)\left(-6-5\right)\right\} \\ =\frac{1}{2}\left\{10\left(2\right)+2\left(9\right)-1\left(-11\right)\right\} \\ =\frac{1}{2}\left\{20+18+11\right\} \\ =\frac{1}{2}\left(49\right) \\ =24.5\mathrm{sq}.\mathrm{units} \end{gathered}$$