Question

Find the area of $\Delta ABC$, in which $A=(2,1),B=(3,4)$ and $C=(-3,-2).$

• A. $3$ square units
• B. $4$ square units
• C. $6$ square units
• D. $12$ square units

Answer 1

Answer: (C)

$6$ square units

Area of triangle with vertices $(x_1,y_1)$, $(x_2,y_2)$ and $(x_3,y_3)$ is:

$A=\frac{1}{2}\left|x_1\right(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2\left)\right|$

Therefore, the area of triangle with vertices $(2,1)$, $(3,4)$ and $(-3,-2)$ is:

$A=\frac{1}{2}\left|2\right(4-(-2))+3(-2-1)-3(1-4\left)\right|$

$=\frac{1}{2}\left|2\right(6)+3(-3)-3(-3\left)\right|$

$=\frac{1}{2}|12-9+9|$

$=\frac{1}{2}\left|12\right|=6$ square units.