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Consider three points $P=\left( -\sin \left( \beta -\alpha \right),-\cos \beta \right),~Q=\left(
\cos \left( \beta -\alpha \right),\sin \beta \right)$ and $R=\left( \cos \left( \beta -\alpha +\theta
\right),\sin \left( \beta -\theta \right) \right)$, where $0<\alpha ,\beta ,\theta <\dfrac{\pi }{4}$. Then
(a) P lies on the line segment RQ
(b) Q lies on the line segment PR
(c) R lies on the line segment QP
(d) P, Q, R are non-collinear

Answer Verified Verified
Hint: Check collinearity of the given 3 points by using section formula.

The given points can be rewritten in simpler terms as
$P=\left( -\sin \left( \beta -\alpha \right),-\cos \beta \right)=\left( {{x}_{1}},{{y}_{1}} \right)\cdots \cdots
\cdots \left( i \right)$
$Q=\left( \cos \left( \beta -\alpha \right),\sin \beta \right)=\left( {{x}_{2}},{{y}_{2}} \right)\cdots \cdots
\cdots \left( ii \right)$
Let the coordinates of the third point $R=\left( \cos \left( \beta -\alpha +\theta \right),\sin \left( \beta -
\theta \right) \right)=\left( {{x}_{3}},{{y}_{3}} \right)$. The ${{x}_{3}}$ coordinate can be simplified as,
${{x}_{3}}=\left( \cos \left( \beta -\alpha +\theta \right) \right)=\cos \left[ \left( \beta -\alpha
\right)+\theta \right]$
Applying the expansion $\cos \left( a+b \right)=\cos a\cos b-\sin a\sin b$,
${{x}_{3}}=\cos \left[ \left( \beta -\alpha \right)+\theta \right]=\cos \left( \beta -\alpha \right)\cos
\theta -\sin \left( \beta -\alpha \right)\sin \theta $
Substituting the corresponding terms from equations $\left( i \right)$ and $\left( ii \right)$,
${{x}_{3}}=\cos \left[ \left( \beta -\alpha \right)+\theta \right]={{x}_{2}}\cos \theta +{{x}_{1}}\sin \theta
$
Now, the ${{y}_{3}}$ coordinate can be simplified as,
${{y}_{3}}=\sin \left( \beta -\theta \right)$
Applying the expansion $\sin \left( a-b \right)=\sin a\cos b-\cos a\sin b$,
${{y}_{3}}=\sin \left( \beta -\theta \right)=\sin \beta \cos \theta -\cos \beta \sin \theta $
Substituting the corresponding terms from equations $\left( i \right)$ and $\left( ii \right)$,
${{y}_{3}}=\sin \left( \beta -\theta \right)={{y}_{2}}\cos \theta +{{y}_{1}}\sin \theta $

So, therefore the third point can be written as,
$R=\left( {{x}_{2}}\cos \theta +{{x}_{1}}\sin \theta ,{{y}_{2}}\cos \theta +{{y}_{1}}\sin \theta \right)\cdots
\cdots \cdots \left( iii \right)$
Consider the line with endpoints PQ. Also consider the point R that lies on the line diving it in the ratio as
below,

Using the section formula, the coordinates of point R can be obtained as,
$R=\left( \dfrac{{{x}_{1}}\cos \theta +{{x}_{2}}\sin \theta }{\sin \theta +\cos \theta },\dfrac{{{y}_{1}}\cos
\theta +{{y}_{2}}\sin \theta }{\sin \theta +\cos \theta } \right)$
From equation $\left( iii \right)$, we have the coordinates of R as $\left( {{x}_{2}}\cos \theta
+{{x}_{1}}\sin \theta ,{{y}_{2}}\cos \theta +{{y}_{1}}\sin \theta \right)$. Comparing this with the above
coordinates, it is clear that the form of the coordinates is not the same.
Therefore, the point R will not lie on the line PQ. It means that the points P, Q and R are not collinear.
Hence, we obtain the correct answer as option (d).

Note: The problem can be solved by applying the condition for collinear points. To check if the points P,
Q and R lie on the same line, consider that point Q lies on line PR. Then, the slope of line PQ and slope of
line QR must be equal for the points to be collinear.
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