Question

If $2x^2-5xy+2y^2=0$ represent two sides of the triangle whose centroid is $(1,1)$ then equation of third side be

• A. $x+y+3=0$
• B. $y-x+3=0$
• C. $x+y-3=0$
• D. $Noneofthese$

Answer 1

Answer: (C)

$x+y-3=0$

Given equation are

$2x^2-5xy+2y^2=0$

$2x^2-(4+1)xy+2y^2=0$

$2x^2-4xy-xy+2y^2=0$

$2x^2(x-2y)-y(x-2y)=0$

$(x-2y)(2x-y)=0$

If $x-2y=0,x=2y......\left(1\right)$

$2x-y=0,y=\frac{x}{2}$

$y=2x$

$\text{let}$ $x=a$

$y=2x=2a$

Point $A(a,2a)$

$\text{If}y=b$

$x=2y$

$x=2b$

Point $B(2b,b)$

Centroid $=(1,1)$ Given that and from (1)

Point $0(0,0)$

So, Centroid formula,

$\frac{a+2b+0}{3}=1,\frac{2a+b+0=1}{3}$

On solving

$a=1,b=1$

so, point $A(1,2)$

$B(2,1)$

From point $A$ and $B$ to.

$y-2=\frac{1-2}{2-1}(x-1)$

$y-2=(-1)(x-1)$

$y-2=-x+1$

$x+y-3=0$

Hence, this is the answer.