The locus of the centroid of a triangle whose | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The centroid $(x, y)$ of a triangle with vertices $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ is given by $x = \frac{x_1 + x_2 + x_3}{3}$ and $y = \frac{

Vedclass · Accepted Answer

$a^2 + b^2 - 1$

Vedclass · Answer

(B) The centroid $(x, y)$ of a triangle with vertices $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ is given by $x = \frac{x_1 + x_2 + x_3}{3}$ and $y = \frac{y_1 + y_2 + y_3}{3}$.Given vertices are $(1, 0)$,$(a \cos t, a \sin t)$,and $(b \sin t, -b \cos t)$.So,$x = \frac{1 + a \cos t + b \sin t}{3} \Rightarrow 3x - 1 = a \cos t + b \sin t$ ...$(i)$And $y = \frac{0 + a \sin t - b \cos t}{3} \Rightarrow 3y = a \sin t - b \cos t$ ...(ii)Squaring and adding equations $(i)$ and (ii):$(3x - 1)^2 + (3y)^2 = (a \cos t + b \sin t)^2 + (a \sin t - b \cos t)^2$$9x^2 - 6x + 1 + 9y^2 = a^2 \cos^2 t + b^2 \sin^2 t + 2ab \sin t \cos t + a^2 \sin^2 t + b^2 \cos^2 t - 2ab \sin t \cos t$$9x^2 + 9y^2 - 6x + 1 = a^2(\cos^2 t + \sin^2 t) + b^2(\sin^2 t + \cos^2 t)$$9x^2 + 9y^2 - 6x + 1 = a^2 + b^2$$9x^2 + 9y^2 - 6x = a^2 + b^2 - 1$Comparing this with the given equation $9x^2 + 9y^2 - 6x = k$,we get $k = a^2 + b^2 - 1$.

The locus of the centroid of a triangle whose vertices are $(1, 0)$,$(a \cos t, a \sin t)$,and $(b \sin t, -b \cos t)$ is $9x^2 + 9y^2 - 6x = k$. Then,the value of $k$ is equal to

Similar Questions

If $A (\cos \alpha, \sin \alpha)$,$B (\sin \alpha, -\cos \alpha)$,and $C (1, 2)$ are the vertices of $\Delta ABC$,find the locus of its centroid as $\alpha$ varies.

If a point $P$ on the line $3x + 5y = 15$ is equidistant from the coordinate axes,then $P$ lies

$A$ variable straight line passes through a fixed point $(a, b)$ intersecting the coordinate axes at $A$ and $B$. If $O$ is the origin,then the locus of the centroid of the triangle $OAB$ is:

If the equation of the locus of a point equidistant from the points $(a_1, b_1)$ and $(a_2, b_2)$ is $(a_1 - a_2)x + (b_1 - b_2)y + c = 0$,then the value of $c$ is

Suppose that the three points $A, B$ and $C$ in the plane are such that their $x$-coordinates as well as $y$-coordinates are in $GP$ with the same common ratio. Then,the points $A, B$ and $C$

Vedclass Test Series

Exam Paper Generator

Online Exam Module