Question

If A(cosa, sina), B(sina, - cosa), C (1,2) are the vertices of a triangle ABC then the locus of its centroid is​

Answer 1

Answer:

Step-by-step explanation:

Given vertices of the triangle are

$\tt{A(cos(\alpha),sin(\alpha)),\,B(sin(\alpha),-cos(\alpha))\,\,\,and\,\,\,C(1,2)}$

Let the coordinates of its centroid be (h,k)

So,

$\sf{h=\dfrac{cos(\alpha)+sin(\alpha)+1}{3}\,\,\,\,\,\,\,\,\,\,and\,\,\,\,\,\,\,\,\,\,k=\dfrac{sin(\alpha)-cos(\alpha)+2}{3}}$

$\sf{\implies\,3h=sin(\alpha)+cos(\alpha)+1\,\,\,\,\,\,\,\,\,\,and\,\,\,\,\,\,\,\,\,\,3k=sin(\alpha)-cos(\alpha)+2}$

$\sf{\implies\,3h-1=sin(\alpha)+cos(\alpha)\,\,\,\,\,\,\,\,\,\,and\,\,\,\,\,\,\,\,\,\,3k-2=sin(\alpha)-cos(\alpha)\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(1)}$

Now, we know,

$\sf{\bullet\,\pink{(a+b)^2+(a-b)^2=2(a^2+b^2)}}$

So,

$\sf{(sin(\alpha)+cos(\alpha))^2+(sin(\alpha)-cos(\alpha))^2=2(sin^2(\alpha)+cos^2(\alpha))}$

$\sf{\implies(3h-1)^2+(3k-2)^2=2(1)}$

$\sf{\implies9h^2-6h+1+9k^2-12k+4=2}$

$\sf{\implies9h^2+9k^2-6h-12k+1+4-2=0}$

$\sf{\implies9h^2+9k^2-6h-12k+3=0}$

$\sf{\implies3h^2+3k^2-2h-4k+1=0}$

$\sf{Hence\,\,,Locus\,\,\,\,is\,\,\,\,3x^2+3y^2-2x-4y+1=0}$