Question

What is the area of the triangle whose vertices are D( 3, 3), E(3, -1), and F(-2, -5)?

Answer 1

Answer-

Area of the triangle is 10 sq.units

Solution-

We know that,

$\text{Area of the triangle}=\dfrac{1}{2}[{x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)]$

Taking,

(x₁, y₁) = (3, 3)

(x₂, y₂) = (3, -1)

(x₃, y₃) = (-2, -5)

Then putting these in the formula,

$\text{Area of the triangle}=\dfrac{1}{2}[3(-1+5)+3(-5-3)-2(3+1)]$

$=\dfrac{1}{2}[3(4)+3(-8)-2(4)]$

$=\dfrac{1}{2}[12-24-8]$

$=\dfrac{1}{2}[-20]$

$=-10$

As area can not be negative, ignoring negative sign,

$\text{Area of the triangle}=10$