UPSC NDA Math Model Paper 9

Given: A (- 3, 1), B (- 1, 4), C (3, 2) and D (1, - 2) are the vertices of a quadilateral ABCD Here, we have to find the area of quadilateral ABCD Area of quadilateral ABCD = Area of ΔABC + Area of Δ ACD Let's find out the area of ΔABC ∵ A (- 3, 1), B (- 1, 4), C (3, 2) are the vertices of ΔABC As we know that, if $A(x_{1}, y_{1}), B(x_{2}, y_{2})$ and $C(x_{3}, y_{3})$ be the vertices of a $\triangle ABC$, then area of $\triangle ABC = \frac{1}{2} \left| \begin{array}{ccc} x_{1} & y_{1} & 1 \\ x_{2} & y_{2} & 1 \\ x_{3} & y_{3} & 1 \end{array} \right|$ $\Rightarrow$ Area of $\triangle ABC = \frac{1}{2} \left| \begin{array}{ccc} -3 & 1 & 1 \\ -1 & 4 & 1 \\ 3 & 2 & 1 \end{array} \right|$ ⇒ Area of Δ ABC = 4 sq. units Similarly, let's find out the area of Δ ACD ∵ A (- 3, 1), C (3, 2) and D (1, - 2) ⇒ Area of $\triangle ACD = \frac{1}{2} \left| \begin{array}{ccc} -3 & 1 & 1 \\ 3 & 2 & 1 \\ 1 & -2 & 1 \end{array} \right|$ ⇒ Area of Δ ACD = 11 sq. units ⇒ Area of quadilateral ABCD = Area of ΔABC + Area of Δ ACD = (8 + 11) sq. units = 19 sq. units Hence, option C is the correct answer.