Question

Find the area of AABC. Where: A = (-3,3), B=(-4,1), C = (-6,0). W Area:

Answer 1

Answer: $1.5\text{ square units}$

Step-by-step explanation:

We know that the area of triangle with coordinates $(x_1,y_1),(x_2,y_2)$ and $(x_3,y_3)$ is given by :-

$\text{Area}=\dfrac{1}{2}|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)|$

Given : The coordinates of ΔABC are  A = (-3,3), B=(-4,1), C = (-6,0).

Then, the area of ΔABC will be :-

$\text{Area}=\dfrac{1}{2}|-3(1-0)+(-4)(0-3)+(-6)(3-1)|\\\\\Rightarrow\text{Area}=\dfrac{1}{2}|-3-4(-3)-6(2)|\\\\\Rightarrow\text{Area}=\dfrac{1}{2}|-3+12-12|\\\\\Rightarrow\text{Area}=\dfrac{1}{2}|-3|=\dfrac{1}{2}(3)=1.5$

Hence, the area of ΔABC= $1.5\text{ square units}$