Question

The vertices of $\Delta$ ABC are $A(5,2),B(4,7)$ and $C(7,-4)$ then find the area.

• A. Area $\Delta$ ABC $=3$ sq. units
• B. Area $\Delta$ ABC $=5$ sq. units
• C. Area $\Delta$ ABC $=8$ sq. units
• D. Area $\Delta$ ABC $=2$ sq. units

Answer 1

The correct option is D Area $\Delta$ ABC $=2$ sq. units

Area of a triangle with vertices $(x_1,y_1)$ ; $(x_2,y_2)$ and $(x_3,y_3)$ is $\left|\frac{x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)}{2}\right|$
Hence, substituting the points $(x_1,y_1)=(5,2)$ ; $(x_2,y_2)=(4,7)$ and $(x_3,y_3)=(7,-4)$ in the area formula, we get
Area of triangle ABC $=\left|\frac{\left(5\right)(7+4)+\left(4\right)(-4-2)+7(2-7)}{2}\right|=\left|\frac{55-24-35}{2}\right|=\frac{4}{2}=2$ sq units.