Question

Find the area of a triangle whose vertices are

(i) (6, 3) (−3, 5) and (4, −2)

(ii) $(a{t}_1^2,2a{t}_1),(a{t}_2^2,2a{t}_2)and(a{t}_3^2,2a{t}_3)$

(iii) (a, c + a), (a, c) and (−a, c − a)

(iv) (1, –1), (–4, 6) and (–3, –5)

Answer 1

We know area of triangle formed by three points is given by

(i) The vertices are given as (6, 3), (−3, 5), (4, −2).

(ii) The vertices are given as .

(iii)

The vertices are given as .

(iv)
Area of the triangle formed by the vertices $\left(x_1,y_1\right),\left(x_2,y_2\right)\mathrm{and}\left(x_3,y_3\right)\mathrm{is}$
$\frac{1}{2}\left|x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_3\left(y_1-y_2\right)\right|$

Now, the given vertices are (1, –1), (–4, 6) and (–3, –5)

Therefore,
$$\begin{gathered} \mathrm{Area}\mathrm{of}\mathrm{triangle}=\frac{1}{2}\left|1\left(6-\left(-5\right)\right)+\left(-4\right)\left(-5-\left(-1\right)\right)+\left(-3\right)\left(-1-6\right)\right| \\ =\frac{1}{2}\left|1\left(6+5\right)+\left(-4\right)\left(-5+1\right)+\left(-3\right)\left(-7\right)\right| \\ =\frac{1}{2}\left|11+\left(-4\right)\left(-4\right)+21\right| \\ =\frac{1}{2}\left|11+16+21\right| \\ =\frac{1}{2}\left|48\right| \\ =24 \end{gathered}$$

Hence, the area of a triangle is 24 square units.