Question

Find the area of triangle having vertices are $A(3,1),B(12,2)$ and $C(0,2)$.

• A. $4$
• B. $6$
• C. $12$
• D. $18$

Answer 1

The correct option is B $6$

The given points are $A(x_1,y_1)=(3,1),B(x_2,y_2)=(12,2)\&C(x_3,y_3)=(0,2)$

Let us obtain $ar\Delta ABC$ by applying the area formula.

$ar\Delta =\frac{1}{2}\{x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)\}$

$\Rightarrow ar\Delta =\frac{1}{2}\{3(2-2)+12(2-1)+0(1-2)\}$ units $=6$ units

$\therefore ar\Delta =6$ units

Hence, option B.