Question

The locus of centroid of triangle with vertices a t a t b t b t cos , sin , sin , cos     and (1,0) where ‘t’ is

parameter is​

Answer 1

Given:

• Vertices of the triangle are $(a\:cost,a\:sint)$, $(b\:sint,\:-b\:cost)$ and $(1,0)$

To find: the locus of the centroid

Formula:

We know that if $(x_{1},y_{1})$, $(x_{2},y_{2})$ and $(x_{3},y_{3})$ be the vertices of a triangle, then the coordinates of its centroid be

$\quad \left(\dfrac{x_{1}+x_{2}+x_{3}}{3},\:\dfrac{y_{1}+y_{2}+y_{3}}{3}\right)$

Solution:

Using the above formula, we find the centroid of the given triangle:

$\quad \left(\dfrac{a\:cost+b\:sint+1}{3},\:\dfrac{a\:sint-b\:cost}{3}\right)$

Let us take:

• $x=\dfrac{a\:cost+b\:sint+1}{3}$

• $\Rightarrow 3x-1=a\:cost+b\:sint$ .....(1)

• and $y=\dfrac{a\:sint-b\:cost}{3}$

• $\Rightarrow 3y=a\:sint-b\:cost$ .....(2)

We can find the locus of the centroid of the given triangle in $x,y$ when the perimeter $'t'$ is removed. This can be done by squaring and adding (1) and (2); we write:

$\quad (3x-1)^{2}+(3y)^{2}\\=a^{2}cos^{2}t+2ab\:sint\:cost+b^{2}sin^{2}t\\+a^{2}sin^{2}t-2ab\:sint\:cost+b^{2}cos^{2}t$

$\Rightarrow (3x-1)^{2}+(3y)^{2}\\=a^{2}(cos^{2}t+sin^{2}t)+b^{2}(sin^{2}t+cos^{2}t)$

$\Rightarrow (3x-1)^{2}+(3y)^{2}=a^{2}+b^{2}$

Answer:

$\quad$ The required locus is given by

$(3x-1)^{2}+(3y)^{2}=a^{2}+b^{2}$