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(B) The area of a triangle with vertices $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ is given by $\Delta = \frac{1}{2} \left| \begin{array}{ccc} x_1 & y_1 &

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(B) The area of a triangle with vertices $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ is given by $\Delta = \frac{1}{2} \left| \begin{array}{ccc} x_1 & y_1 & 1 \ x_2 & y_2 & 1 \ x_3 & y_3 & 1 \end{array} ight|$.Given the determinant $\left|\begin{array}{ccc}2 a & x_{1} & y_{1} \ 2 b & x_{2} & y_{2} \ 2 c & x_{3} & y_{3}\end{array}ight|=\frac{a b c}{2}$.Taking $2$ common from the first column,we get $2 \left|\begin{array}{ccc} a & x_{1} & y_{1} \ b & x_{2} & y_{2} \ c & x_{3} & y_{3}\end{array}ight|=\frac{a b c}{2}$.Dividing by $abc$,we get $2 \left|\begin{array}{ccc} 1 & x_{1}/a & y_{1}/a \ 1 & x_{2}/b & y_{2}/b \ 1 & x_{3}/c & y_{3}/c \end{array}ight|=\frac{1}{2}$.Thus,$\left|\begin{array}{ccc} 1 & x_{1}/a & y_{1}/a \ 1 & x_{2}/b & y_{2}/b \ 1 & x_{3}/c & y_{3}/c \end{array}ight|=\frac{1}{4}$.The area of the triangle is $\frac{1}{2} \left| \begin{array}{ccc} x_1/a & y_1/a & 1 \ x_2/b & y_2/b & 1 \ x_3/c & y_3/c & 1 \end{array} ight|$.Since the determinant value is $\frac{1}{4}$,the area is $\frac{1}{2} 	imes \frac{1}{4} = \frac{1}{8}$.

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