Coordinate Geometry Question 16
Question: The points $ (1,1) $ , $ (0,{{\sec }^{2}}\theta ),(cose{c^{2}}\theta ,0) $ are collinear for
[Roorkee 1963]
Options:
A) $ \theta =\frac{n\pi }{2} $
B) $ \theta \ne \frac{n\pi }{2} $
C) $ \theta =n\pi $
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
The given points are collinear, if Area $ =\frac{1}{2}\begin{vmatrix} 1 & 1 & 1 \\ 0 & {{\sec }^{2}}\theta & 1 \\ cose{c^{2}}\theta & 0 & 1 \\ \end{vmatrix} =0 $
$ \Rightarrow 1({{\sec }^{2}}\theta )+1(cose{c^{2}}\theta )-1(cose{c^{2}}\theta .{{\sec }^{2}}\theta )=0 $
$ \Rightarrow \frac{1}{{{\cos }^{2}}\theta }+\frac{1}{{{\sin }^{2}}\theta }-\frac{1}{{{\sin }^{2}}\theta {{\cos }^{2}}\theta }=0 $
$ \Rightarrow \frac{1}{{{\cos }^{2}}\theta {{\sin }^{2}}\theta }-\frac{1}{{{\sin }^{2}}\theta {{\cos }^{2}}\theta }=0\Rightarrow 0=0 $ Therefore, the points are collinear for all values of q, except only $ \theta =\frac{n\pi }{2} $ because at $ \theta =\frac{n\pi }{2},{{\sec }^{2}}\theta =\infty . $
BETA
