The area of a triangle whose vertices are $A (3, 7), B (1, 5)$ and $C (6, 11)$ is

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Solution

The correct option is B $1$
$Area of Δ ABC = \frac{1}{2} ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ \begin{matrix} 3 & 7 & 1 1 & 5 & 1 6 & 11 & 1 \end{matrix} ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ R_{1} \to R_{1} - R_{2}, R_{2} \to R_{2} - R_{3} = \frac{1}{2} ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ \begin{matrix} 2 & 2 & 0 - 5 & - 6 & 0 6 & 11 & 1 \end{matrix} ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ (expanding the determinant through C_{3}) = \frac{1}{2} | 1 (- 12 + 10) | = \frac{1}{2} | 2 | = 1 sq. unit$

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The area of a triangle whose vertices are A(3,7),B(1,5) and C(6,11) is

The correct option is B 1 Area of ΔABC=12∣∣ ∣∣∣∣ ∣∣3711516111∣∣ ∣∣∣∣ ∣∣R1→R1−R2, R2→R2−R3=12∣∣ ∣∣∣∣ ∣∣220−5−606111∣∣ ∣∣∣∣ ∣∣(expanding the determinant through C3)=12|1(−12+10)|=12|2|=1 sq. unit

The area of a triangle whose vertices are $A (3, 7), B (1, 5)$ and $C (6, 11)$ is