Question

the centroid of a triangle formed by the points( 0,0) (cos teta ,sin teta ) and( sin teta ,_cos teta,) lies on the line y =2×, then teta=​

Answer 1

Question :-

The centroid of a triangle formed by the points (0,0), (cosθ, sinθ) and (sinθ, - cosθ) lies on the line y =2x, then θ = ___

$\large\underline{\sf{Solution-}}$

The coordinates of triangle are

$\rm \: (0,0), \: (cos\theta , \: sin\theta ) \: and \: (sin\theta , \: - \: cos\theta )$

Let assume that centroid of a triangle be (h, k).

We know,

Centroid of a triangle is defined as the point at which the medians of the triangle meet and is represented by the symbol G.

Let A(x₁, y₁), B(x₂, y₂) and C(x₃, y₃) be the vertices of a triangle and G(x, y) be the centroid of the triangle, then the coordinates of G is given by

$\begin{gathered}\boxed{\tt{ (x, y) = \bigg(\dfrac{x_{1}+x_{2}+x_{3}}{3}, \dfrac{y_{1}+y_{2}+y_{3}}{3}\bigg)}} \\ \end{gathered}$

So, using this result, we get

$\rm \: (h, \: k) = \bigg(\dfrac{0 + cos\theta + sin\theta }{3}, \: \dfrac{0 + sin\theta - cos\theta }{3} \bigg)$

$\rm \: (h, \: k) = \bigg(\dfrac{cos\theta + sin\theta }{3}, \: \dfrac{sin\theta - cos\theta }{3} \bigg)$

So, it means Coordinates of Centroid of a triangle is

$\boxed{\rm{  \:\rm \: (h, \: k) = \bigg(\dfrac{cos\theta + sin\theta }{3}, \: \dfrac{sin\theta - cos\theta }{3} \bigg) \: \: }}$

Now, it is given that

The centroid of the triangle lies on the line y = 2x.

It means, Centroid (h, k) lies on the line y = 2x

$\rm\implies \:k \: = \: 2h$

On substituting the values of h and k, we get

$\rm \: \dfrac{sin\theta - cos\theta }{3} = 2 \times \dfrac{cos\theta + sin\theta }{3}$

$\rm \: sin\theta - cos\theta =2 cos\theta + 2sin\theta$

$\rm \: 2cos\theta + cos\theta = - 2sin\theta + sin\theta$

$\rm \: 3cos\theta = - sin\theta$

$\rm \: tan\theta \: = \: - \: 3$

$\rm \: \theta \: = \: {tan}^{ - 1} ( - \: 3)$

$\rm \: \theta \: = \: - \: {tan}^{ - 1} ( 3)$

Hence,

$\rm\implies \:\boxed{\rm{  \:\rm \: \theta \: = \: - \: {tan}^{ - 1} ( 3) \: \: }}$

$\rule{190pt}{2pt}$

Additional Information :-

$\begin{gathered}\boxed{\begin{array}{c|c} \bf Function & \bf Range \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf y =  {sin}^{ - 1}(sinx) & \sf  x \:  \: if -\dfrac{\pi  }{2} \leqslant x \leqslant \dfrac{\pi  }{2}\\ \\ \sf y =  {cos}^{ - 1}(cosx) & \sf x \:  \: if \: 0 \leqslant y \leqslant \pi \\ \\ \sf y =  {tan}^{ - 1}(tanx) & \sf x \:  \: if \:  - \dfrac{\pi  }{2} < x < \dfrac{\pi  }{2}\\ \\ \sf y =  {cosec}^{ - 1}(cosecx) & \sf x \:  \: if \: x \:  \in \: \bigg[ - \dfrac{\pi}{2}, \: \dfrac{\pi  }{2}\bigg] -  \{0 \}\\ \\ \sf y =  {sec}^{ - 1}(secx) & \sf x \:  \: if \: x \:  \in \: [0, \: \pi] \:   -  \: \bigg\{\dfrac{\pi  }{2}\bigg\}\\ \\ \sf y =  {cot}^{ - 1}(cotx) & \sf x \:  \: if \:  \:  \in \: \bigg( -  \dfrac{\pi  }{2} , \dfrac{\pi  }{2}\bigg) -  \{0 \} \end{array}} \\ \end{gathered}$

Answer 2

PROVIDED INFORMATION :-

• the centroid of a triangle formed by the points( 0,0) (cos teta ,sin teta )

GIVEN :-

• vertices of triangle are O0(0, 0), 4(cos 0, sin 0) and B (sine, - cos)

TO FIND :-

• lies on the line y =2×, then teta = ?

SOLUTION :-

Given, vertices of triangle are O0(0, 0),

4(cos 0, sin 0) and B (sine, - cose), then

Coordinates of centroid = ( cos θ + sin θ ) / 3 , sin θ - cos θ /3 )

Since, centroid lies on the line y=2x

sin θ - cos θ /3 = 2 cos θ + 2 sin θ /3

sin θ = - 3 cos θ = θ = tan - ¹ (-3)