# 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

367.2k+ views

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.

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.

Last updated date: 29th Sep 2023

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