Answer

Verified

462k+ 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.

Recently Updated Pages

When people say No pun intended what does that mea class 8 english CBSE

Name the states which share their boundary with Indias class 9 social science CBSE

Give an account of the Northern Plains of India class 9 social science CBSE

Change the following sentences into negative and interrogative class 10 english CBSE

Advantages and disadvantages of science

10 examples of friction in our daily life

Trending doubts

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

Which are the Top 10 Largest Countries of the World?

Give 10 examples for herbs , shrubs , climbers , creepers

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

10 examples of law on inertia in our daily life

Write a letter to the principal requesting him to grant class 10 english CBSE

Difference Between Plant Cell and Animal Cell

Change the following sentences into negative and interrogative class 10 english CBSE