Question

The points $\left( 1,1 \right),\left( 0,{{\sec }^{2}}\theta \right),\left( \cos e{{c}^{2}}\theta ,0 \right)$ are collinear for
(a) $\theta =\dfrac{n\pi }{2}$
(b) $\theta \ne \dfrac{n\pi }{2}$
(c) $\theta =n\pi$
(d) None of these

Answer 1

Hint:We will use the concept in which if three collinear points are given then they will never form any triangle. So, because of this we will use the formula of the area of triangle which is given by $\dfrac{1}{2}\left| \begin{matrix} {{x}_{1}} & {{y}_{1}} & 1 \\\ {{x}_{2}} & {{y}_{2}} & 1 \\\ {{x}_{3}} & \,{{y}_{3}} & 1 \\\ \end{matrix} \right|=0$. We have equated the area equal to zero. This is due to the fact that since the collinear points do not form a triangle so their area of a triangle is zero. We will use this formula in order to solve the question.

Complete step-by-step answer:
Here we use the trick which says that the area of the triangle can also be carried out to find whether the points are collinear. By above mentioned condition which is already given is that the points are collinear. This means that the area of the triangle is zero. So now we will use the formula of area of triangle $\dfrac{1}{2}\left| \begin{matrix} {{x}_{1}} & {{y}_{1}} & 1 \\\ {{x}_{2}} & {{y}_{2}} & 1 \\\ {{x}_{3}} & \,{{y}_{3}} & 1 \\\ \end{matrix} \right|=0$. Here, we are taking the area equal to zero because the points are collinear and they will form no triangle resulting in a zero area. Therefore, we have

& \dfrac{1}{2}\left| \begin{matrix} 1 & 1 & 1 \\\ 0 & {{\sec }^{2}}\theta & 1 \\\ \cos e{{c}^{2}}\theta & 0 & 1 \\\ \end{matrix} \right|=0 \\\ & \Rightarrow \dfrac{1}{2}\left[ 1\left( {{\sec }^{2}}\theta \right)-1\left( -\cos e{{c}^{2}}\theta \right)+1\left( 0-{{\sec }^{2}}\theta \cos e{{c}^{2}}\theta \right) \right]=0 \\\ & \Rightarrow \dfrac{1}{2}\left[ {{\sec }^{2}}\theta +\cos e{{c}^{2}}\theta -{{\sec }^{2}}\theta \cos e{{c}^{2}}\theta \right]=0 \\\ & \Rightarrow {{\sec }^{2}}\theta +\cos e{{c}^{2}}\theta -{{\sec }^{2}}\theta \cos e{{c}^{2}}\theta =0 \\\ \end{aligned}$$ Now convert it into $\sin $ and $\cos $ terms we get $${{\sec }^{2}}\theta +\cos e{{c}^{2}}\theta -{{\sec }^{2}}\theta \cos e{{c}^{2}}\theta =0$$. $$\begin{aligned} & \Rightarrow \dfrac{1}{{{\sin }^{2}}\theta }+\dfrac{1}{{{\cos }^{2}}\theta }-\dfrac{1}{{{\sin }^{2}}\theta {{\cos }^{2}}\theta }=0 \\\ & \Rightarrow \dfrac{{{\sin }^{2}}\theta +{{\cos }^{2}}\theta }{{{\sin }^{2}}\theta {{\cos }^{2}}\theta }-\dfrac{1}{{{\sin }^{2}}\theta {{\cos }^{2}}\theta }=0 \\\ & \Rightarrow \dfrac{{{\sin }^{2}}\theta +{{\cos }^{2}}\theta -1}{{{\sin }^{2}}\theta {{\cos }^{2}}\theta }=0 \\\ \end{aligned}$$ Now, it is noticeable that $${{\sin }^{2}}\theta {{\cos }^{2}}\theta \ne 0$$ and this implies that $${{\sin }^{2}}\theta {{\cos }^{2}}\theta \ne 0..........(ii)$$. By using formula ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$ implies $${{\sin }^{2}}\theta \left( 1-{{\sin }^{2}}\theta \right)\ne 0$$. So now split the expression and we get $${{\sin }^{2}}\theta \ne 0$$ or $$\left( 1-{{\sin }^{2}}\theta \right)\ne 0$$. If $${{\sin }^{2}}\theta \ne 0$$ then $$\sin \theta \ne \sin 0$$ means $\theta \ne n\pi $ also if we consider $$\left( 1-{{\sin }^{2}}\theta \right)\ne 0$$ then $$\begin{aligned} & 1-{{\sin }^{2}}\theta \ne 0 \\\ & \Rightarrow {{\sin }^{2}}\theta \ne 1 \\\ & \Rightarrow \sin \theta \ne \sin \dfrac{\pi }{2} \\\ \end{aligned}$$ This means that $$\theta \ne \dfrac{n\pi }{2}$$. Hence the correct option is (b). Note: We can check the co linearity of the points, if asked, by finding the slope. This is done below. The points are collinear if the slope of any pairs of two points out of three points are equal. Here it means that if we consider $\left( 1,1 \right)$ as point A $\left( 0,{{\sec }^{2}}\theta \right)$ as B point and $\left( \cos e{{c}^{2}}\theta ,0 \right)$ as C point, also if we take the pairs AB, BC and CA together then Slope of AB = Slope of BC = Slope of CA. Now, we find the slope of AB first with the points A $\left( 1,1 \right)$ and B $\left( 0,{{\sec }^{2}}\theta \right)$ and the formula for finding slope is given by $m=\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}...(i)$ so, after substituting points A $\left( 1,1 \right)$ and B$\left( 0,{{\sec }^{2}}\theta \right)$ in $(i)$ we get $m=\dfrac{{{\sec }^{2}}\left( \theta \right)-1}{0-1}...(ii)$. Now, we apply identity ${{\tan }^{2}}\theta ={{\sec }^{2}}\theta -1$ in $(ii)$ we have $\dfrac{{{\sec }^{2}}\left( \theta \right)-1}{0-1}=\dfrac{{{\tan }^{2}}\theta }{-1}$ which is actually $-{{\tan }^{2}}\theta $. So Slope of AB = Slope of BC = Slope of CA = $-{{\tan }^{2}}\theta $. As all the slopes are equal therefore the points are collinear. The area of the triangle which is considered by most students is $\dfrac{1}{2}$ of product of base and height. But here we are given three collinear points for which we have to use the formula, $$\dfrac{1}{2}\left| \begin{matrix} {{x}_{1}} & {{y}_{1}} & 1 \\\ {{x}_{2}} & {{y}_{2}} & 1 \\\ {{x}_{3}} & \,{{y}_{3}} & 1 \\\ \end{matrix} \right|=0$$ moreover we can also substitute the basic trigonometric identity which is given by ${{\sin }^{2}}\theta =1-{{\cos }^{2}}\theta $ in $(ii)$ and solve. While solving the question we should never forget to write the second term which is here ${{y}_{1}}$ term in negative and then solve further. One has to focus during the calculations otherwise, it can lead to wrong multiplication and it will lead to wrong answers.