Question

find the area of triangle whose vertices are A(2,3) B (-1,0)and C (2,-4)​

Answer 1

Given: Points:

• A(2, 3)
• B(-1, 0)
• C(2, -4)

To find: The area of the triangle so formed.

Answer:

Area of a triangle:

$\tt \dfrac{1}{2}\ \times\ \bigg[x_1\bigg(y_2\ -\ y_3\bigg)\ +\ x_2\bigg(y_3\ -\ y_1\bigg)\ +\ x_3\bigg(y_1\ -\ y_2\bigg)\bigg]$

From the given points,

$\tt x_1\ =\ 2\\\\x_2\ =\ -1\\\\x_3\ =\ 2\\\\y_1\ =\ 3\\\\y_2\ =\ 0\\\\y_3\ =\ -4$

Using these values in the formula,

$\tt \implies\ \dfrac{1}{2}\ \times\ \bigg[2\bigg(0\ +\ 4\bigg)\ +\ -1\bigg(-4\ -\ 3\bigg)\ +\ 2\bigg(3\ -\ 0\bigg)\bigg]\\\\\\=\ \ \ \ \ \ \dfrac{1}{2}\ \times\ \bigg[8\ +\ 7\ +\ 6\bigg]\\\\\\=\ \ \ \ \ \ \dfrac{1}{2}\ \times\ 21\\ \\\\=\ \ \ \ \ 10.5\ units$

Therefore, the area of the triangle is 10.5 units.