UPSC NDA Math Model Paper 5

Here, we have to find the value of a for which the points (- 8, 4), (- 2, 4) and (5, a) are collinear Let A = (- 8, 4), B = (- 2, 4) and C = (5, a) Let $x_{1}=-8, y_{1}=4, x_{2}=-2, y_{2}=4, x_{3}=5$ and $y_{3}=a$ 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 $\Delta ABC$ then area of Δ ABC = | A | where A = $\frac{1}{2} \begin{vmatrix} x_{1} & y_{1} & 1 \\ x_{2} & y_{2} & 1 \\ x_{3} & y_{3} & 1 \end{vmatrix}$ $\Rightarrow A = \frac{1}{2} \begin{vmatrix} -8 & 4 & 1 \\ -2 & 4 & 1 \\ 5 & a & 1 \end{vmatrix}$ $\Rightarrow A = 3a - 12$ ∵ 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. $\Rightarrow A = 3a - 12 = 0$ $\Rightarrow k = 4$ Hence, option A is the correct answer.