The points (1, 1),(0, sec^2 theta),and (csc^2 | vedclass

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(B) The points $(x_1, y_1)$,$(x_2, y_2)$,and $(x_3, y_3)$ are collinear if the area of the triangle formed by them is $0$.Area $= \frac{1}{2} |x_1(y_2

Vedclass · Accepted Answer

$	heta 
eq \frac{n\pi}{2}$

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(B) The points $(x_1, y_1)$,$(x_2, y_2)$,and $(x_3, y_3)$ are collinear if the area of the triangle formed by them is $0$.Area $= \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)| = 0$Substituting the given points $(1, 1)$,$(0, \sec^2 	heta)$,and $(\csc^2 	heta, 0)$:$\frac{1}{2} |1(\sec^2 	heta - 0) + 0(0 - 1) + \csc^2 	heta(1 - \sec^2 	heta)| = 0$$\sec^2 	heta + \csc^2 	heta(1 - \sec^2 	heta) = 0$$\sec^2 	heta + \csc^2 	heta - \csc^2 	heta \sec^2 	heta = 0$$\frac{1}{\cos^2 	heta} + \frac{1}{\sin^2 	heta} - \frac{1}{\sin^2 	heta \cos^2 	heta} = 0$$\frac{\sin^2 	heta + \cos^2 	heta}{\sin^2 	heta \cos^2 	heta} - \frac{1}{\sin^2 	heta \cos^2 	heta} = 0$$\frac{1}{\sin^2 	heta \cos^2 	heta} - \frac{1}{\sin^2 	heta \cos^2 	heta} = 0$This identity $0 = 0$ holds for all $	heta$ where $\sec 	heta$ and $\csc 	heta$ are defined.Since $\sec 	heta$ is undefined at $	heta = (2n+1)\frac{\pi}{2}$ and $\csc 	heta$ is undefined at $	heta = n\pi$,the points are collinear for all $	heta$ except $	heta = \frac{n\pi}{2}$.

The points $(1, 1)$,$(0, \sec^2 \theta)$,and $(\csc^2 \theta, 0)$ are collinear for:

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