Q191 / 270 · NDA 2021 (Phase 2) · Mathematics

The coordinates of three consecutive vertices of a parallelogram $A B C D$ are $A(1,3), B(-1,2)$ and $C(3,5)$. What is the area of the parallelogram?

  • A
    1 square unit
  • B
    $\frac{3}{2}$ square units
  • Correct answer
    C
    2 square units
  • D
    $\frac{5}{2}$ square units

Solution

$\begin{aligned} & \operatorname{Ar} \Delta \mathrm{ABC}=\frac{1}{2}\left|x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_3\left(y_1-y_2\right)\right| \\ & \quad=\frac{1}{2}|1(2-5)-1(5-3)+3(3-2)|=1 \\ & \operatorname{Ar} \Delta \mathrm{ACD}=\frac{1}{2}|1(5-6)+3(6-3)+5(3-5)|=1 \end{aligned}$ $\therefore \operatorname{Ar} A B C D=1+1=2$ square units.

Hence, the correct answer is option: 'C'.

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