UPSC NDA I 2021 Mathematics Solved Paper

CALCULATION: Given: $(p, p-3),(q+3, q)$ and $(6,3)$ are three points in a $2D$ plane. Statement 1: The points lie on a straight line. As we know that, if three points $A(x_{1}, y_{1}), B(x_{2}, y_{2})$ and $C(x_{3}, y_{3})$ aresaid to be collinear if the area of $\Delta ABC$ is zero i.e $\begin{vmatrix} x_{1} & y_{1} & 1 \\ x_{2} & y_{2} & 1 \\ x_{3} & y_{3} & 1 \end{vmatrix}=0$ Let $A=(p, p-3), B=(q+3, q)$ and $C=(6,3)$ $\Rightarrow \begin{vmatrix} p & p-3 & 1 \\ q+3 & q & 1 \\ q+3 & 3 & 1 \end{vmatrix}$ $=p(q-3)-(p-3)(q+3-6)+1(3q+9-6q)$ $\Rightarrow \begin{vmatrix} p & p-3 & 1 \\ q+3 & q & 1 \\ q+3 & 3 & 1 \end{vmatrix}=0$ So, as we can see that area of triangle $ABC=0$ for any value of $p$ and $q$. Hence, statement 1 is true. Statement 2: The points alwayslie in the first quadrant only for any value of $p$ and $q.$ As we can see that, for any value of $p$ and $q$ it is not necessary that the points lies in the first quadrant only. Hence, statement 2 is false. Hence, the correct option is $1.$