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Question: If the centroid of the triangle formed by the points \(\left( {0,0} \right)\), \(\left( {\cos \theta...

If the centroid of the triangle formed by the points (0,0)\left( {0,0} \right), (cosθ,sinθ)\left( {\cos \theta ,\sin \theta } \right) and (sinθ,cosθ)\left( {\sin \theta , - \cos \theta } \right) lies on the line y=2xy = 2x , then θ\theta is equal to
A.tan12{\tan ^{ - 1}}2
B.tan13{\tan ^{ - 1}}3
C.tan1(3){\tan ^{ - 1}}\left( { - 3} \right)
D.tan1(2){\tan ^{ - 1}}\left( { - 2} \right)

Explanation

Solution

Here we are given that the centroid is formed by the points (0,0)\left( {0,0} \right), (cosθ,sinθ)\left( {\cos \theta ,\sin \theta } \right) and (sinθ,cosθ)\left( {\sin \theta , - \cos \theta } \right)
So, we need to apply these vertices in the formula to find the centroid. Then it is given that the centroid lies on the line y=2xy = 2x. Thus, we shall apply the x-coordinate and the y-coordinate of the centroid on the line and we need to simplify the obtained equation to find the desired answer.
Formula to be used:
a) The formula to calculate the centroid of a given triangle is as follows.
c=(x1+x2+x33,y1+y2+y33)c = \left( {\dfrac{{{x_1} + {x_2} + {x_3}}}{3},\dfrac{{{y_1} + {y_2} + {y_3}}}{3}} \right)
Here cc is the centroid of the triangle; x1{x_1} , x2{x_2} and x3{x_3} are the x-coordinates of the given three vertices and y1{y_1} , y2{y_2} and y3{y_3} are the y-coordinates of the three vertices.
b) sinθcosθ=tanθ\dfrac{{\sin \theta }}{{\cos \theta }} = \tan \theta

Complete answer:

We are given that the centroid of the triangle formed by the points (0,0)\left( {0,0} \right), (cosθ,sinθ)\left( {\cos \theta ,\sin \theta } \right) and (sinθ,cosθ)\left( {\sin \theta , - \cos \theta } \right)
We need to calculate the angle.
Thus, the given three vertices are (0,0)\left( {0,0} \right), (cosθ,sinθ)\left( {\cos \theta ,\sin \theta } \right) and (sinθ,cosθ)\left( {\sin \theta , - \cos \theta } \right)
We all know that the centroid of the triangle is c=(x1+x2+x33,y1+y2+y33)c = \left( {\dfrac{{{x_1} + {x_2} + {x_3}}}{3},\dfrac{{{y_1} + {y_2} + {y_3}}}{3}} \right)
Now, we need to apply the x-coordinates and the y-coordinates of the three vertices in the above formula.
Thus, we get c=(0+cosθ+sinθ3,0+sinθcosθ3)c = \left( {\dfrac{{0 + \cos \theta + \sin \theta }}{3},\dfrac{{0 + \sin \theta - \cos \theta }}{3}} \right)
c=(cosθ+sinθ3,sinθcosθ3)\Rightarrow c = \left( {\dfrac{{\cos \theta + \sin \theta }}{3},\dfrac{{\sin \theta - \cos \theta }}{3}} \right)
Thus, we have x-coordinate as cosθ+sinθ3\dfrac{{\cos \theta + \sin \theta }}{3} and y-coordinate as sinθcosθ3\dfrac{{\sin \theta - \cos \theta }}{3}
Also, we are given that the centroid of the triangle lies on the line y=2xy = 2x
So, we need to apply x=cosθ+sinθ3x = \dfrac{{\cos \theta + \sin \theta }}{3} and y=sinθcosθ3y = \dfrac{{\sin \theta - \cos \theta }}{3} on the line y=2xy = 2x
Thus, we have sinθcosθ3=2cosθ+sinθ3\dfrac{{\sin \theta - \cos \theta }}{3} = 2\dfrac{{\cos \theta + \sin \theta }}{3}
Now, we need to simplify the above equation to find the angle.
Hence, sinθcosθ=2cosθ+2sinθ\sin \theta - \cos \theta = 2\cos \theta + 2\sin \theta
sinθ2sinθ=2cosθ+cosθ\Rightarrow \sin \theta - 2\sin \theta = 2\cos \theta + \cos \theta
sinθ=3cosθ\Rightarrow - \sin \theta = 3\cos \theta
3=sinθcosθ\Rightarrow - 3 = \dfrac{{\sin \theta }}{{\cos \theta }}
3=tanθ\Rightarrow - 3 = \tan \theta (Here we applied sinθcosθ=tanθ\dfrac{{\sin \theta }}{{\cos \theta }} = \tan \theta )
θ=tan1(3)\Rightarrow \theta = {\tan ^{ - 1}}\left( { - 3} \right)

Therefore, we found that θ=tan1(3)\theta = {\tan ^{ - 1}}\left( { - 3} \right) and option C) is the answer.

Note:
The centroid is the center point of a triangle and we can also say that the centroid is the point of intersection of three medians of the triangle where median is a line segment that joins a vertex to the midpoint of the opposite side. Also, the centroid is formed by the three vertices of the triangle.