UPSC NDA Math Model Paper 9

Here, we have to find the value of t for which the points (2, 4), (2t, 6t) and (3, 8) are collinear Let A = (2, 4), B = (2t, 6t) and C = (3, 8) Let $x_{1}=2, y_{1}=4, x_{2}=2 t, y_{2}=6 t, x_{3}=3$ and $y_{3}=8$ As we know that, if $A(x_{1}, y_{1}), B(x_{2}, y_{2})$ and $C(x_{3}, y_{3})$ are the vertices of a Δ ABC then area of Δ ABC = |A| where $A=\frac{1}{2} \begin{vmatrix} x_{1} & y_{1} & 1 \\ x_{2} & y_{2} & 1 \\ x_{3} & y_{3} & 1 \end{vmatrix}$ $\Rightarrow A=\frac{1}{2} \begin{vmatrix} 2 & 4 & 1 \\ 2t & 6t & 1 \\ 3 & 8 & 1 \end{vmatrix}$ ⇒ |A| = t - 2 ∵ The given points are collinear. As we know that, if the points $A(x_1, y_1), B(x_2, y_2)$ and $C(x_3, y_3)$ are collinear then area of ΔABC = 0. ⇒ A = t - 2 = 0 ⇒ k = 2 Hence, option B is the correct answer.